P= S pi.

If the gas is collected above the liquid, then when calculating its partial pressure, it should be borne in mind that it is equal to the difference between the total pressure and the partial pressure of the vapor of the liquid. For example, for a gas collected over water,

The law of equivalents. The equivalent of a substance is that amount of it that combines with 1 mole of hydrogen atoms or replaces the same number of hydrogen atoms in chemical reactions. Equivalent mass E is the mass of one equivalent of a substance. The equivalent volume of a gas is the volume occupied under given conditions by one equivalent of a substance. The equivalent (equivalent mass) can be calculated from the composition of the compound of a given element with any other, the equivalent (equivalent mass) of which is known, according to the law of equivalents: the masses of interacting substances A + B ® C + D are proportional to their equivalent masses:

Based on the law of equivalents, you can calculate the equivalent mass of a substance:

where M is the molar mass of the element, oxide, acid, base or salt, g/mol; Z- the degree of oxidation of the element in the reaction product, the product of the number of atoms of the element and the degree of oxidation of the element in oxides, the basicity of the acid, the acidity of the base, the product of the number of metal atoms and the degree of oxidation of the metal in the salt.

Example1. Determine the mass fraction of aluminum in its oxide and calculate how much aluminum can theoretically be isolated from 15 tons of bauxite with an Al 2 O 3 content of 87%.

Solution. Find the molar mass of Al 2 O 3:

Let's take the amount of substance Al 2 O 3 equal to 1 mol, then the amount of aluminum substance will be 2 mol. The mass of aluminum oxide will be 102 g, and the mass of aluminum 2 × 27 \u003d 54 g. Let's calculate the mass fraction of aluminum in its oxide:

.

Let's calculate the mass of pure Al 2 O 3 in bauxite and the mass of aluminum, which can be obtained from 15 tons of bauxite:

Example2. When calcining 10 g of a certain substance, 6.436 g of CuO and 3.564 g of CO 2 were obtained. Write the compound formula.

Solution. 1. Find the amount of copper (II) oxide substance:

1 mol of CuO contains 1 mol of Cu and O, therefore n(Cu)== n(O, CuO) = 0.081 mol.

2. Find the amount of carbon monoxide substance (IV):

1 mol of CO 2 contains 1 mol of C and 2 mol of O, therefore n(C) = 0.081 mol, = 2 x 0.081 = 0.162 mol.

3. The total amount of oxygen substance n(O) = 0.081 + + 0.162 = 0.243 mol.

4. Let's write the ratio of the amount of substance of the elements:

n(cu) : n(C) : n(O) = 0.081:0.081:0.243 = 1:1:(0.243/0.081) = 1:1:3.

The resulting integers are the stoichiometric indices of the formula of the substance. Therefore, the chemical formula of the desired substance is CuCO 3.

Example 3 The compound of sulfur with fluorine contains 62.8% sulfur and 37.2% fluorine. This compound with a volume of 118 ml in a gaseous state (temperature 7 ° C, pressure 96.34 kPa) has a mass of 0.51 g. What is the true formula of the compound?

Solution. 1. Calculate the true molar mass of the compound according to the Mendeleev-Clapeyron equation:

2. Let x and y is the number of sulfur and fluorine atoms, respectively, in the S molecule x F y. Knowing the percentage of each element in the compound and its molar mass, we calculate

3. Thus, the simplest formula of the compound SF, and its molar mass M = 32 + 19 = 51 g/mol. Since the ratio of the true and simplest molar masses , then the desired formula contains 2 times more atoms of each type. Hence, the formula of the compound is S 2 F 2.

Example 4 . Oxidation of 2.81 g of cadmium yielded 3.21 g of cadmium oxide. Calculate the equivalent mass of cadmium and determine its valency.

Solution. 1. By the mass of cadmium and the mass of its oxide, we find the mass of oxygen: m(O)= m(CdO)- m(Cd) = 3.21 - 2.81 = 0.4 g.

2. The formation of cadmium oxide can be written as a reaction scheme Cd + O ® CdO, for which we compose a proportion according to the law of equivalents:

3. Comparing the numerical values ​​of the equivalent mass and the molar mass of cadmium, we find . Therefore, the valency of cadmium is 2.

Example 5 . Manganese (IV) oxide loses oxygen upon calcination, forming Mn 3 O 4 . What volume of oxygen at a temperature of 27 ° C and a pressure of 1.1 atm will be released from 0.58 kg of MnO 2?

Solution. 1. Let's write the decomposition reaction equation

from which it follows that 3 mol of MnO 2 give 1 mol of oxygen.

Find the amount of substance MnO 2:

therefore, it is formed

.

2. Given that 1 atm = 101325 Pa, according to the Mendeleev-Clapeyron equation, we get

Example 6. To a solution containing 0.2 mol of ferric chloride (FeCl 3), 0.24 mol of sodium hydroxide was added. How much iron hydroxide did this result in?

Solution. From the reaction equation

FeCl 3 + 3 NaOH ® Fe(OH) 3 + 3 NaCl

it follows that 1 mol of FeCl 3 interacts with 3 mol of NaOH. Therefore, for the reaction with 0.2 mol of ferric chloride, 0.2 × 3 = 0.6 mol of sodium hydroxide is required.

According to the condition of the problem, the amount of NaOH substance is 0.24 mol, i.e. he is in short supply. Further calculation is carried out for sodium hydroxide. Let's make a proportion:

3 mol NaOH - 1 mol FeCl 3

0.24 mol NaOH - X mol FeCl 3,

of which the amount of iron (III) hydroxide substance

Exercise. Solve problems.

1. The composition of the hematite mineral is expressed by the ratio m(fe) : m(O)=7:3. How many grams of iron can be obtained from 50 g of this mineral?

2. On an industrial scale, cadmium oxide is obtained by burning cadmium in excess dry air. Determine the quantitative composition of cadmium oxide and derive its formula if the combustion of 2.1 g of cadmium produces 2.4 g of oxide.

3. Cryolite has the composition AlF 3 ∙3NaF. Calculate the mass fraction of aluminum fluoride in cryolite.

5. To analyze copper chloride and determine its quantitative composition, a solution of silver nitrate was poured into a solution containing 0.4 g of copper chloride. A precipitate of silver chloride was formed weighing 0.849 g. Determine the quantitative composition and derive the formula of copper chloride.

6. After preliminary purification of bauxite, an anhydrous product was obtained, consisting mainly of aluminum oxide and containing 0.3% silicon (IV) oxide and 0.048% iron (III) oxide. What is the percentage of silicon and iron in this product?

7. How much manganese can be isolated by aluminothermy from 20 kg of pyrolusite containing 87% manganese (IV) oxide?

8. Give the chemical name of the mineral and calculate the mass fraction of chlorine in carnallite KMgCl 3 ∙6H 2 O.

10. How much concentrate with a copper content of 60% can be obtained from 1 ton of ore containing 3% chalcocite (Cu 2 S) and 2% covellite (CuS)?

11. Give the chemical name of the mineral and calculate the percentage of copper in chrysocolla CuSiO 3 ∙2H 2 O.

12. What mass of iron can be obtained from 2 tons of iron ore containing 94% Fe 3 O 4 .

13. What mass of aluminum can be obtained from 1 ton of nepheline NaAlSiO 4?

14. Compose the formula of iron (III) dihydroxosulfate and calculate the percentage of sulfur oxide (VI) in it.

15. The KHSO 4 compound can be thought of as being composed of K 2 O and SO 3 . Find the percentage of sulfur oxide (VI) in this compound and name it.

16. Write the formula for iron (III) sulfate and calculate the iron content in this compound.

17. Determine how much silver and silver oxide can be obtained from 10 kg of silver chloride.

18. Calculate the content of copper (II) oxide and name the compound (CuOH) 2 CO 3 .

19. Give a chemical name to the compound FeCl 3 6H 2 O and calculate the percentage of chlorine.

20. Name the compound (NiOH) 3 (PO 4) and calculate the percentage of nickel in it.

21. The substance consists of sulfur and carbon. To determine its quantitative composition, 0.3045 g of this substance was taken. All the sulfur contained in the sample taken was converted into barium sulfate, the mass of which is 1.867 g. Find the quantitative composition of the substance and indicate its formula.

22. The substance consists of aluminum and chlorine. From a certain amount of substance, 1.7196 g of AgCl and 0.2038 g of Al 2 O 3 were obtained. Find the quantitative composition and establish the formula of the substance.

23. When reducing 2.4 g of copper oxide with hydrogen, 0.54 g of H 2 O was obtained. Find the quantitative composition and write the formula of the oxide.

24. When heated, Bertolet's salt decomposes into oxygen and potassium chloride. Calculate the quantitative composition of Bertolet's salt and derive its formula if 0.62 g of KCl is obtained from the decomposition of 1.02 g of salt.

25. The substance consists of potassium, sulfur and oxygen. Sulfur and oxygen contained in 0.871 g of this substance were isolated in the form of BaSO 4 weighing 1.167 g. Find the quantitative composition and establish the formula of the substance.

26. When decomposing a certain amount of a substance consisting of copper, carbon, oxygen and hydrogen, 1.432 g of CuO, 0.396 g of CO 2 and 0.159 g of water were obtained. Find the quantitative composition and formula of the substance.

27. The substance consists of copper and sulfur. From 0.667 g of this substance, 0.556 g of CuO was obtained. Calculate the percentage composition and write down the formula of the substance.

28. When a solution of silver nitrate was added to a solution of 0.408 g of copper chloride, a precipitate of silver chloride weighed 0.86 g. Calculate the quantitative composition of chloride and establish its formula.

29. When analyzing a sample of iron ore weighing 125 g, 58 g of magnetite Fe 3 O 4 were found in it. Calculate the mass fraction of iron in the ore sample.

30. Compose the true formula of a compound containing 1.59% hydrogen, 22.21% nitrogen and oxygen. The molar mass of the compound is 63 g/mol.

31. Determine the true formula of a compound containing 3.03% hydrogen, 31.62% phosphorus and oxygen. The molar mass of the compound is 80 g/mol.

32. What is the true formula of a compound containing 6.75% hydrogen, 39.97% carbon and oxygen. The relative vapor density of this substance in terms of carbon dioxide is 4.091.

33. During the combustion of 10.5 liters of organic matter, 16.8 liters of carbon monoxide (IV) reduced to normal conditions and 13.5 g of water were obtained. The density of this substance is 1.875 g/cm3. Derive the formula for this substance.

34. Determine the chemical formula of a substance that contains five mass parts of calcium and three mass parts of carbon.

35. The substance consists of 32.8% Na, 12.9% Al, 54.3% F. Write the formula of the substance.

36. Find the simplest formula of a substance consisting of carbon, hydrogen, sulfur, mercury and chlorine, based on the following data: a) when 3.61 g of a substance is oxidized, 1.72 g of carbon monoxide (IV) and 0.90 g of water are obtained; b) 0.467 g of barium sulfate was obtained from 0.722 g of the substance; c) 0.859 g of silver chloride was obtained from 1.0851 g of the substance.

37. During the firing of pyrite, a gas is released containing 40% sulfur and 60% oxygen and having an air density at n.o. 2.76. Set the gas formula.

38. Qualitative analysis showed that malachite consists of copper, carbon, oxygen and hydrogen. When decomposing a certain amount of malachite, 0.48 g of copper (II) oxide, 0.132 g of carbon monoxide (IV) and 0.053 g of water were obtained. Derive the formula for malachite.

39. Potassium alum contains 8.23% potassium, 5.7% aluminum, 13.5% sulfur, 27.0% oxygen and 45.5% water. What is the formula for alum?

40. When receiving steel, impurities of sulfur and phosphorus are especially undesirable. Phosphorus in steel is contained in the form of an oxygen compound containing 43.66% phosphorus and 56.34% oxygen. The density of this compound in air under normal conditions is 4.9. Derive the formula of the given oxygen compound of phosphorus.

41. Ore containing 696 tons of magnetic iron ore was delivered to the plant. From this ore, 504 tons of iron were smelted. Write down the formula of magnetic iron ore, if it is known that it consists only of iron and oxygen.

42. Find the formula of barium chloride crystalline hydrate, knowing that 36.6 g of salt lose 5.4 g in mass when calcined.

43. Find the simplest formula of a substance containing (by mass) 43.4% sodium, 11.3% carbon and 45.3% oxygen.

44. The substance contains (by mass) 40.21% potassium, 26.80% chromium and 32.99% oxygen. Find its simplest formula.

45. The compound contains 46.15% carbon. The rest is nitrogen. The air density is 1.79. find the true formula of the compound.

46. ​​With the complete combustion of 2.66 g of a certain substance, 1.54 g of CO 2 and 4.48 g of SO 2 were obtained. Find the simplest formula of a substance.

47. Find the molecular formula of the compound of boron with hydrogen, if the mass of 1 liter of this gas is equal to the mass of 1 liter of nitrogen, and the boron content in the substance is 78.2%.

48. The compound of sulfur with fluorine contains 62.8% S and 37.2% F. The volume of this compound in the form of a gas is 118 ml, at 7 ° C and 98.64 kPa its mass is 0.51 g. What is the true formula of the compound?

49. Find the formula of a substance containing 85.71% C and 14.29% H, if the density of this gas in air is 4.83.

50. With the complete combustion of organic matter weighing 13.8 g, 26.4 g of carbon monoxide (IV) and 16.2 g of water were obtained. find the molecular formula of a substance if its hydrogen vapor density is 23.

51. The chemical compound consists (by mass) of 25.48% copper, 12.82% sulfur, 25.64% oxygen and 36.06% water. Find the formula of the compound and name it.

52. Determine the formula of a gaseous substance containing (by mass) 20% hydrogen and 80% carbon, if its hydrogen density is 15.

53. With the complete combustion of 0.23 g of a substance consisting of carbon, hydrogen and oxygen, 0.27 g of water and 224 ml of carbon dioxide were obtained (the volume of gas was measured under normal conditions). Determine the molecular formula of a substance if its vapor density in air is 1.59.

54. The composition of the compound includes carbon, hydrogen and nitrogen. Carbon is 79.12% in it. The mass of nitrogen obtained from 0.546 g of the compound is 0.084 g. The molar mass of the substance is 182. Derive its formula.

55. Determine the formula of a crystalline hydrate containing 8.11% Al, 28.83% O, 14.41% S and 48.65% H 2 O.

56. What is the formula of a substance containing 42.9% SiO 2 and 57.1% MgO?

57. Determine the formula of a crystalline hydrate containing 16.08% Na, 4.2% C, 16.78% O and 62.94% H 2 O.

58. Set the formula of a crystalline hydrate containing 16.08% Na, 11.94% S, 23.89% O and 47% H 2 O.

59. Calculate the molar mass of benzene if 1.1 l of its vapor at 91 ° C and 81313 Pa has a mass of 2.31 g.

60. The mass of 584 ml of gas at 21 ° C and normal pressure is 1.44 g. Calculate the molar mass of the gas.

61. The mass of 0.36 l of the vapor of a substance at 98 ° C and 98.642 kPa is 1.8 g. Calculate the molar mass of the substance.

62. The mass of 454 ml of gas at 44 ° C and 97309 Pa is 1.19 g. Calculate the molar mass of the gas.

63. Calculate the mass of 1 m 3 of air at 37 ° C and 83200 Pa.

64. Calculate the volume that it occupies at 27 ° C and 760 mm Hg. Art. 1 kg of air.

65. A cylinder with a capacity of 20 liters contains 3 kg of oxygen. Calculate the pressure in the balloon at 20°C.

66. Calculate at what pressure 5 kg of nitrogen will occupy a volume of 50 liters if the temperature is 500 ° C?

67. A cylinder with a capacity of 10 liters at 27 ° C contains 3 × 10 23 oxygen molecules. Calculate the pressure of oxygen in the cylinder.

68. A flask with a capacity of 0.75 liters filled with oxygen at 20 ° C has a mass of 132 g. The mass of an empty flask is 130.79 g. Calculate the oxygen pressure in the flask.

69. A steel cylinder for storing compressed gases contains 64 kg of oxygen. Determine the mass of carbon dioxide, which is filled with the same balloon under the same conditions.

70. Some gas was collected in a closed cylinder with a volume of 41 liters at a temperature of 627 ° C and a pressure of 1.2 atm. The mass of the gas in the cylinder is 42.7 g. Find the molar mass of the gas and determine what kind of gas it is if sulfur is included in its composition.

71. For analysis at 25 °C and 779 mm Hg. Art. the gas sample was taken into a 100 ml flask. The mass of the flask with gas is 16.392 g, the mass of the empty flask is 16.124 g. Determine the molar mass of the gas.

72. A flask with a capacity of 232 ml was filled with some gas at a temperature of 17 ° C and a pressure of 752 mm Hg. Art. The mass of the flask increased by 0.27 g. Calculate the molar mass of the gas.

73. To analyze the composition of the gas, a gasometer with a capacity of 20 liters was filled at a pressure of 1.025 atm and a temperature of 17 ° C. The mass of the gasometer has increased by 10 g. Calculate the molar mass of the gas.

74. A cylinder with a capacity of 1 liter was filled with gas at a temperature of 21 ° C and a pressure of 1.05 atm. The mass of the gas in the cylinder is 1.48 g. Calculate the molar mass of the gas.

75. Determine how many molecules are contained in 3 liters of some gas at a pressure of 1520 mm Hg. Art. and a temperature of 127 °C.

76. Determine at what temperature is 0.2 g of some gas, occupying a volume of 0.32 liters, if the gas pressure is 1.5 atm, and its density in air is 1.52.

77. What is the temperature of a gas if its pressure is 30 atm, mass 1.5 kg, volume 170 l, air density 1.08?

78. At a pressure of 98.7 kPa and a temperature of 91 ° C, the gas occupies a volume of 680 ml. Find the volume of gas under normal conditions.

79. The cylinder contains gas at a temperature of 27 °C. Determine what part of the gas will remain in the cylinder if, with the cylinder open, the gas temperature is increased by 100 ° C.

80. Gas pressure in a closed vessel at 12 ° C is 100 kPa. What will be the pressure of the gas if the vessel is heated to 303 K?

81. The volume of 0.111 g of some substance is 26 ml at 17 ° C and 104 kPa. Calculate the molar mass of the gas.

82. At -23 ° C, the volume of gas is 8 liters. At what temperature will the volume of a gas be 10 liters if the pressure is left unchanged?

83. In a closed cylinder with a capacity of 40 liters is 77 g of CO 2. The pressure gauge connected to the cylinder shows a pressure of 106.6 kPa. Calculate the temperature of the gas in the cylinder.

84. At 27 ° C, the volume of gas is 600 ml. What volume will the gas occupy when the temperature is increased by 30 K, if the pressure is left unchanged.

85. Find the mass of 1 m 3 of air at 17 ° C and a pressure of 624 mm Hg. Art.

86. Gas at 10 ° C and a pressure of 960 hPa occupies a volume of 50 ml. At what pressure will a gas occupy a volume of 10 ml if its temperature rises by 10 K?

87. Determine the molar mass of organic matter, knowing that 0.39 g of its vapor at a temperature of 87 ° C and a pressure of 936 mm Hg. Art. occupy a volume of 120 ml.

88. Calculate the mass of 3 m 3 of oxygen at a temperature of 27 ° C and a pressure of 780 mm Hg. Art.

89. Calculate the mass of oxygen that filled the gasometer with a capacity of 14.5 liters at a temperature of 17 ° C and a pressure of 16 atm.

90. Determine the molar mass of a gas, 0.96 g of which occupies a volume of 0.41 liters at a temperature of 27 ° C and a pressure of 1.2 atm.

91. A vessel with a capacity of 5 liters contains 7 g of nitrogen at 273 K. Determine the pressure of the gas. At what temperature will it become equal to 1 atm?

92. A vessel with a capacity of 15 liters contains 21 g of nitrogen at 400 K. Determine the pressure of the gas.

93. How much does 1 liter of gas weigh under normal conditions, if its density in air is 1.52?

94. A vessel with a capacity of 15 liters contains 21 g of nitrogen at 273 K. Determine the pressure of the gas.

95. A liter of some gas weighs 2.86 g under normal conditions. Determine the molar mass of the gas and its density in air.

96. 2.8 liters of gas weigh 2 g under normal conditions. Determine the molar mass of the gas and its density in air.

97. Determine the mass of 190 ml of benzene vapor at a temperature of 97 ° C and a pressure of 740 mm Hg. Art.

98. What volume is occupied by 4.2 g of nitrogen at a temperature of 16 ° C and a pressure of 771 mm Hg. Art.?

99. Calculate the molar mass of an unknown gas and its density in air, knowing that the mass of 0.5 liters of this gas under normal conditions is 0.5804 g.

100. Determine the molar mass of ether, knowing that 312 ml of its vapor at a temperature of 47 ° C and a pressure of 800 mm Hg. Art. weigh 0.925 g.

101. Find the mass of 1 liter of air at a temperature of 40 ° C and a pressure of 939 mm Hg. Art.

102. Determine the molar mass of a substance if the mass of 312 ml of its vapor at a temperature of 40 ° C and a pressure of 939 mm Hg. Art. equal to 1.79 g.

103. 52.5 g of nitrogen occupy a volume of 41 liters at a temperature of 7 ° C. determine the gas pressure.

104. Determine the molar mass of a gas, 0.96 g of which occupies a volume of 0.41 liters at a temperature of 27 ° C and a pressure of 1.2 atm.

105. Calculate the mass of oxygen that filled the gasometer with a capacity of 14.5 liters at a temperature of 17 ° C and a pressure of 16 atm.

106. In a closed vessel with a capacity of 3 liters, 0.5 liters of nitrogen and 2.5 liters of hydrogen are mixed. Their initial pressure is 103.5 and 93.7 kPa, respectively. Determine the partial pressures of gases and the total pressure of the mixture.

107. Mixed 2 liters of carbon dioxide (= 1 atm) and 5.6 liters of nitrogen (= 96.9 kPa). What are the partial pressures of the gases in the mixture and what is its total pressure?

108. Calculate the volume fractions (in percent) of neon and argon in the mixture if their partial pressures are 203.4 and 24.6 kPa, respectively.

109. Calculate the volume fractions (in percent) of carbon oxides (II) and (IV), the partial pressure of which is 0.24 and 0.17 kPa, respectively.

110. The total pressure of a mixture of argon and hydrogen is 108.6 kPa. What is the volume fraction of argon if the partial pressure of hydrogen is 105.2 kPa?

111. In a vessel with a capacity of 6 liters, there is nitrogen under a pressure of 3 × 10 6 Pa. After the addition of oxygen, the pressure of the mixture increased to 3.4×10 6 Pa. What is the volume fraction of oxygen in the mixture?

112. In a gas tank above water at a temperature of 25 ° C there is 5.2 liters of oxygen at a pressure of 102.4 kPa. What is the volume of dry oxygen if the pressure of saturated water vapor at the same temperature is 3.164 kPa?

113. As a result of the reaction of 4.45 g of metal with hydrogen, 5.1 g of hydride was formed. Determine the equivalent mass of the metal.

114. When 0.385 g of metal interacted with chlorine, 1.12 g of chloride of this metal was formed. Calculate the equivalent mass of the given metal.

115. The reaction of 0.44 g of metal with bromine required 3.91 g of bromine. Determine the equivalent mass of the metal.

116. Determine the equivalent mass of a divalent metal and name it if 0.26 liters of oxygen, measured under normal conditions, were required for the complete combustion of 3.2 g of metal.

117. When hydrogen sulfide was passed through a solution containing 7.32 g of divalent metal chloride, 6.133 g of its sulfide was obtained. Determine the equivalent mass of the metal.

118. During the decomposition of 4.932 g of metal oxide, 0.25 liters of oxygen were obtained, reduced to normal conditions. Determine the equivalent mass of the metal.

119. When a metal plate weighing 10.2 g interacted with a solution of copper (II) sulfate, the mass of the plate increased by 1.41 g. Calculate the equivalent mass of the metal.

120. Lead oxide contains 7.14% (by mass) of oxygen. Determine the equivalent mass of lead.

121. The compound of a metal with a halogen contains 64.5% (by mass) of halogen, the oxide of the same metal contains 15.4% (by mass) of oxygen. Determine the equivalent mass of halogen and name it.

122. The reduction of 6.33 g of metal oxide consumed 0.636 liters of hydrogen, reduced to normal conditions. Determine the equivalent mass of the metal.

123. Calculate the equivalent mass of metal, 2 g of which are combined with 1.39 g of sulfur or 6.95 g of bromine.

124. It has been established that 0.321 g of aluminum and 1.168 g of zinc displace the same amount of hydrogen from acid. Find the equivalent weight of zinc if the equivalent weight of aluminum is 8.99 g/eq.

125. How many liters of hydrogen, reduced to normal conditions, will be required to reduce 112 g of metal oxide containing 71.43% of the metal? What is the equivalent mass of metal?

126. Calculate the molar and equivalent mass of a divalent metal if 2.2 g of it is displaced from an acid by 0.81 liters of hydrogen at 22 ° C and 102.9 kPa. name metal.

127. Calculate the equivalent mass of acid if it took 28.9 ml of sodium hydroxide solution with a concentration of 0.1 mol / l to neutralize 0.234 g of it.

128. It took 3.04 g of hydrochloric acid to neutralize 2 g of base. Calculate the equivalent mass of the base.

129. Calculate the equivalent of phosphoric acid in the reactions:

K 2 CO 3 + 2 H 3 PO 4 ® 2 KH 2 PO 4 + CO 2 + H 2 O;

K 2 CO 3 + H 3 PO 4 ® K 2 HPO 4 + CO 2 + H 2 O;

3 K 2 CO 3 + 2 H 3 PO 4 ® 2 K 3 PO 4 + 3CO 2 + 3 H 2 O.

130. Calculate the equivalent of potassium carbonate in the reactions

K 2 CO 3 + HI ® KHCO 3 + KI;

K 2 CO 3 + 2 HI ® H 2 CO 3 + 2 KI.

131. In technology, copper oxide is obtained by calcining copper with a lack of air. Determine the equivalent mass of copper if 9 g of copper oxide is obtained when calcining 8 g of copper.

132. The mineral chalcocite (copper sheen) contains 20% sulfur. Determine the equivalent mass of the metal and the formula of chalcosine.

133. One of the ways to obtain metals is the reduction of their oxides with hydrogen. Calculate the equivalent mass of the metal if it is known that the reduction of 3.4 g of metal oxide required as much hydrogen as it is released during the reaction of 6.54 g of zinc with acid.

134. Calculate the equivalent mass of the metal if 8.61 g of silver chloride was obtained from 4.93 g of metal chloride by reaction with silver nitrate.

135. 0.58 g of copper was dissolved in nitric acid. The resulting salt was calcined to give 0.726 g of copper oxide. Calculate the equivalent mass of copper.

136. 1.02 g of metal was dissolved in acid. In this case, 0.94 liters of hydrogen, measured under normal conditions, were released. Calculate the equivalent mass of the metal.

137. One of the operations in the production of steel by the Bessemer method is the combination of basic metal oxides with silicon oxide (IV) according to the equation MnO + SiO 2 → MnSiO 3. When using 100 g of slag containing 25% silicon (IV) oxide, the equivalent mass of which is 15 g/mol, 109.2 g of manganese silicate was formed. Calculate the equivalent mass of manganese silicate.

138. To restore 15.9 g of iron chloride, 2.8 liters of hydrogen were used, brought to normal conditions. Calculate the equivalent mass of ferric chloride.

139. When burning 5.0 g of metal, 9.44 g of metal oxide is formed. Determine the equivalent mass of the metal.

140. It has been established that 1.0 g of a certain metal combines with 8.89 g of bromine or with 1.78 g of sulfur. Find the equivalent masses of bromine and metal, knowing that the equivalent mass of sulfur is 16.0 g/eq.

X: y \u003d 1.32 / 1.32: 1.98 / 1.32 \u003d 1: 1.5, and then multiply both values ​​\u200b\u200bof the last ratio by two: x: y \u003d 2: 3. Thus, the simplest oxide formula chromium Cr2O3. EXAMPLE 9 Complete combustion of a substance weighing 2.66 g produced CO2 and SO2 with masses of 1.54 g and 4.48 g, respectively. Find the simplest formula of a substance. Solution The composition of the combustion products shows that the substance contained carbon and sulfur. In addition to these two elements, oxygen could also be included in its composition. The mass of carbon that was part of the substance can be found from the mass of CO2 formed. The molar mass of CO2 is 44 g/mol, while 1 mole of CO2 contains 12 g of carbon. Let's find the mass of carbon m contained in 1.54 g of CO2: 44/12 = 1.54/m; m \u003d 12 1.54 / 44 \u003d 0.42 g. Calculating in a similar way the mass of sulfur contained in 4.48 g of SO2, we obtain 2.24 g. Since the mass of sulfur and carbon is 2.66 g, this substance does not contain oxygen and the formula of the substance CxSy: x: y = 0.42/12: 2.24/32 = 0.035: 0.070 = 1: 2. Therefore, the simplest formula of the substance is CS2. To find the molecular formula of a substance, it is necessary, in addition to the composition of the substance, to know its molecular weight. PRI me R 10 Gaseous connection of nitrogen with hydrogen contains 12.5% ​​(wt.) hydrogen. The density of the compound by hydrogen is 16. Find the molecular formula of the compound. Solution The desired formula of the substance NxHy: x: y \u003d 87.5 / 14: 12.5 / 1 \u003d 6.25: 12.5 \u003d 1: 2. The simplest formula of the compound NH2. This formula corresponds to a molecular weight equal to 16 a.m.u. We find the true molecular weight of the compound based on its hydrogen density: M = 2 16 = 32 a.m.u. Therefore, the formula of the substance is N2H4. PRI mme R 11 When calcining hydrated zinc sulfate weighing 2.87 g, its mass decreased by 1.26 g. Set the formula of crystalline hydrate. Solution When calcined, the crystalline hydrate decomposes: t ZnSO 4 nH2O → ZnSO4 + nH2O М(ZnSO4) = 161 g/mol; M(H2O) = 18 g/mol. It follows from the condition of the problem that the mass of water is 1.26 g, and the mass of ZnSO4 is (2.87-1.26) = 1.61 g. Then the amount of ZnSO4 will be: 1.61/161 = = 0.01 mol, and number of moles of water 1.26/18 = 0.07 mol. Therefore, for 1 mole of ZnSO4 there are 7 moles of H2O and the formula of ZnSO4 crystalline hydrate is 7H2O Example 12 Find the mass of sulfuric acid required to completely neutralize sodium hydroxide weighing 20 g. . М(H2SO4) = 98 g/mol; M(NaOH) = 40 g/mol. By condition: ν(NaOH) = 20/40 = 0.5 mol. According to the reaction equation, 1 mol of H2SO4 reacts with 2 mols of NaOH, 0.25 mol of H2SO4 or 0.25 98 = 24.5 g reacts with 0.5 mol of NaOH. sawdust weighing 1.76 g; resulting in a mixture of metal chlorides weighing 4.60 g. Calculate the mass of copper that reacted. Solution The reactions proceed according to the schemes: 1) Cu + Cl2 = CuCl2 2) 2Fe + 3Cl2 = 2FeCl3 М(Cu) = 64 g/mol; M(Fe) = 56 g/mol; М(CuCl2) = 135 g/mol; M (FeCl3) \u003d 162.5 g / mol. Let us denote the copper content in the mixture through x g. Then the iron content in the mixture will be (1.76 - x) g. From equations (1.2) it follows that the mass of the formed copper (II) chloride "a" will be g, the mass of iron (III) chloride "b" will be b = (1.76 - x) 162.5 / 56 g. According to the condition of the problem, the mass of the mixture of copper (II) and iron (III) chlorides, i.e. a + b \u003d 4.60 g. Hence 135x / 64 + 162.5 (1.76 - x) / 56 \u003d 4.60. Therefore, x = 0.63, i.e. the mass of copper is 0.63 g. .48 dm3 (N.O.). Calculate the composition of the initial mixture (ω, %). Solution Reaction equations: 1) KHCO3 + HCl = KCl + H2O + CO2 2) KOH + HCl = KCl + H2O M(KHCO3) = 100 g/mol; M(KCl) = 74.5 g/mol; M(KOH) = 56 g/mol. According to the condition of the problem, the volume of gas (CO2) according to reaction (1) is 4.48 dm3 or 0.2 mol. Then it follows from the reaction equation (1) that the initial amount in a mixture of potassium bicarbonate is 0.2 mol or 0.2 100 \u003d 20 g and the same amount is formed 0.2 mol KCl or 0.2 74.5 \u003d 14.9 d. Knowing the total mass of KCl formed as a result of reactions (1 and 2), it is possible to determine the mass of KCl formed by reaction (2). It will be 22.35 - 14.9 \u003d 7.45 g or 7.45 / 74.5 \u003d 0.1 mol. The formation of 0.1 mol KCl according to reaction (2) will require the same amount of KOH, that is, 0.1 mol or 0.1 56 = 5.60 g. Therefore, the content of the initial components in the mixture will be: 5.6 100 /25.6 = 21.9% KOH and 20.0 100/25.6 = 78.1% KHCO3. Problems 51 During the decomposition of metal(II) carbonate weighing 21.0 g, CO2 was released with a volume of 5.6 dm3 (n.o.). Set the salt formula. 52 Find the formulas of compounds with composition in mass fractions of a percent: a) sulfur - 40 and oxygen - 60; b) iron - 70 and oxygen - 30; c) chromium - 68.4 and oxygen - 31.6; d) potassium - 44.9; sulfur - 18.4 and oxygen - 36.7; e) hydrogen - 13.05; oxygen - 34.78 and carbon - 52.17; f) magnesium - 21.83; phosphorus - 27.85; and oxygen - 50.32. 53 Determine the formulas of compounds that have a composition in mass fractions of a percent: a) potassium - 26.53; chromium - 35.35 and oxygen - 38.12; b) zinc - 47.8 and chlorine - 52.2; c) silver - 63.53; nitrogen - 8.24 and oxygen - 28.23; d) carbon - 93.7; hydrogen - 6.3. 54 Determine the simplest formulas of minerals that have a composition in mass fractions of a percent: a) copper - 34.6; iron - 30.4; sulfur - 35.0; b) calcium - 29.4; sulfur - 23.5; oxygen - 47.1; c) calcium - 40.0; carbon - 12.0; oxygen - 48.0; d) sodium - 32.9; aluminum - 12.9; fluorine - 54.2. 55 Set the formulas: a) vanadium oxide, if the oxide weighing 2.73 g contains a metal weighing 1.53 g: b) mercury oxide, if complete decomposition of its mass 27 g releases oxygen with a volume of 1.4 dm3 (n.o.) ? 56 Establish the formula of a substance consisting of carbon, hydrogen and oxygen in a mass ratio of 6: 1: 8, respectively, if its vapor density in air is 2.07. 57 Determine the formula of a compound having a composition in mass fractions of a percent of metal - 38.71; phosphorus - 20.00; oxygen - 41.29. 58 Find the formula of a compound with a molar mass of 63 g / mol, having a composition in mass fractions of a percent: hydrogen - 1.59; nitrogen - 22.21 and oxygen - 76.20. 59 Establish the formula of the compound (M = 142 g/mol) having the composition in mass fractions of a percentage: sulfur - 22.55; oxygen - 45.02 and sodium - 32.43. 60 Find the formula of the compound (M = 84 g / mol), having a composition in mass fractions of a percent: magnesium - 28.5; carbon - 14.3; oxygen - 57.2. 61 Find the formula of a compound (M = 136 g/mol) having a composition in mass fractions of a percent: calcium - 29.40; hydrogen - 0.74; phosphorus - 22.80; oxygen - 47.06. 62 Establish the formula of the compound (M = 102 g/mol) having the composition in mass fractions of a percent: aluminum - 52.9; oxygen - 47.1. 63 When a substance weighing 3.4 g was burned in oxygen, nitrogen and water were formed, weighing 2.8 g and 5.4 g, respectively. Set the formula of the substance. 64. Find the formula of the compound (M = 310 g / mol), which has a composition in mass fractions of a percent: calcium - 38.75; phosphorus - 20.00; oxygen - 41.25. 65 Find the formula of a hydrocarbon having a composition in mass fractions of a percent: carbon - 82.76; hydrogen - 17.24. In a gaseous state, a hydrocarbon with a volume of 1.12 dm3 (n.o.) has a mass of 2.9 g. 66 Find the formula for a compound of boron with hydrogen (borane), which has a composition in mass fractions of a percent: boron - 78.2; hydrogen - 21.8; if the mass of 1 dm3 of this gas is equal to the mass of 1 dm3 of nitrogen (n.c.). 67 Find the formula of a substance that has a composition in mass fractions of a percent: carbon - 93.75; hydrogen - 6.25. The density of this substance in air is 4.41. 68 Find the formula of a substance if its hydrogen density is 49.5; and the composition is expressed in mass fractions of a percent: carbon - 12.12; oxygen - 16.16; chlorine - 71.72. 69 The combustion of a hydrocarbon weighing 4.3 g produced carbon dioxide weighing 13.2 g. The vapor density of a hydrocarbon with respect to hydrogen is 43. What is the formula of a hydrocarbon? 70 Complete combustion of a sulfur compound with hydrogen produces water and sulfur dioxide with masses of 3.6 g and 12.8 g, respectively. Set the formula of the starting substance. 71 What is the formula of silane (silane), if it is known that when it is burned with a mass of 6.2 g, silicon dioxide is formed with a mass of 12.0 g? The density of silicon hydrogen in air is 2.14. 72 Complete combustion of organic matter weighing 13.8 g produced carbon dioxide and water with masses of 26.4 g and 16.2 g, respectively. The hydrogen vapor density of this substance is 23. Determine the formula of the substance. 73 Combustion of an unknown substance weighing 5.4 g in oxygen produced nitrogen, carbon dioxide and water weighing 2.8 g; 8.8 g; 1.8 g respectively. Determine the formula of a substance if its molar mass is 27 g/mol. 74 Mass fractions of sodium, calcium and silicon (IV) oxides in window glass are 13.0, respectively; 11.7 and 75.3%. What is the molar ratio of these oxides to express the composition of the glass? 75 Establish the formula of sodium sulfate crystalline hydrate if the weight loss on ignition is 55.91% of the mass of crystalline hydrate. 76 Establish the formula of barium chloride crystalline hydrate, if the weight loss was 5.4 g upon calcination of a salt weighing 36.6 g. 26 g. 78 The residue after calcination of copper(II) sulfate crystal hydrate weighing 25 g was 16 g. Determine the formula of the crystal hydrate. 79 During dehydration of the crystalline hydrate of copper (II) chloride weighing 1.197 g, the loss in mass was 0.252 g. Establish the formula of the crystalline hydrate. 80 Find the formula of calcium chloride crystalline hydrate, if 1.44 g of water was released when it was calcined with a mass of 5.88 g. The composition of potassium alum includes water of crystallization with a mass fraction of 45.5%. Calculate how many moles of water per mole of KAl(SO4)2. 83 Determine the formula of crystalline hydrate, in which the mass fractions of the elements are: magnesium - 9.8%; sulfur - 13.0%; oxygen - 26.0%; water - 51.2%. 84 Set the formula of crystalline hydrate, the composition of which is expressed in mass fractions of a percent: iron - 20.14; sulfur - 11.51; oxygen - 63.35; hydrogen - 5.00. 85 Find the formula of crystalline soda, which has a composition in mass fractions of a percent: sodium - 16.08; carbon - 4.20; oxygen - 72.72; hydrogen - 7.00. 86 Establish the formula of calcium sulfate crystalline hydrate if the mass loss was 0.36 g upon calcination of the crystalline hydrate weighing 1.72 g. 94 g. Set the formula of this crystalline hydrate. 88 Determine the formula of carnallite xKCl yMgCl2 zH2O, if it is known that when calcining 5.55 g, its mass decreased by 2.16 g; and on calcination of the precipitate obtained by the action of an alkali solution on a solution containing the same amount of salt, the losses amount to 0.36 g. 89 The composition of the compound includes carbon, hydrogen, chlorine, and sulfur. The combustion of this substance with a mass of 1.59 g produces carbon dioxide and water with masses of 1.76 g and 0.72 g, respectively. When this substance with a mass of 0.477 g is dissolved and silver nitrate is added to a solution, a precipitate with a mass of 0.861 g is formed. The molar mass of the substance is 159 g / mol. Set the formula of the substance. 90 Determine the formula of double iron(III) sulfate and ammonium if it is known that when it is dissolved with a mass of 19.28 g in water and then an excess of concentrated NaOH solution is added, a gas with a volume of 896 cm3 (N.O.) is released and a brown precipitate is formed , upon calcination of which the mass of the residue is 3.20 g. 91 Determine the formula of the compound in which the mass fractions of the elements are: metal - 28%; sulfur - 24%; oxygen - 48%. 92 Natural crystalline hydrate contains water of crystallization and salt with mass fractions of 56% and 44%, respectively. Derive the formula of the crystalline hydrate, if it is known that the salt, which is part of the crystalline hydrate, colors the flame yellow and forms a white precipitate insoluble in water and acids with a solution of barium chloride. 93 Calculate the volume of hydrogen (N.O.) that will be released during the interaction of aluminum weighing 2.7 g with a solution containing KOH weighing 20 g. 94 When metal (II) with a mass of 6.85 g dm3 (n.o.s.). Define metal. 95 To a solution containing iron (III) sulfate weighing 40 g was added a solution containing NaOH weighing 24 g. What is the mass of the precipitate formed? 96 What mass of calcium carbonate should be taken in order to fill a balloon with a capacity of 40 dm3 at 188 K and a pressure of 101.3 kPa with carbon dioxide obtained by its decomposition? 97 Bertolet's salt decomposes on heating to form potassium chloride and oxygen. What volume of oxygen at 0 °C and a pressure of 101325 Pa can be obtained from one mole of Bertolet salt? 98 Determine the mass of salt formed during the interaction of calcium oxide weighing 14 g with a solution containing nitric acid weighing 35 g. 99 To a solution containing calcium chloride weighing 0.22 g was added a solution containing silver nitrate weighing 2.00 d. What is the mass of the precipitate formed? What substances will be in solution? 100 Under the action of hydrochloric acid on an unknown metal weighing 22.40 g, metal(II) chloride is formed and a gas is released with a volume of 8.96 dm3 (n.o.). Identify the unknown metal. 101 Calculate the content of impurities in mass fractions of a percent in limestone, if carbon dioxide with a volume of 20 dm 102 What mass of aluminum will be required to obtain hydrogen necessary for the reduction of copper oxide (II), obtained by thermal decomposition of malachite weighing 6.66 g? 103 The reduction of an oxide of an unknown metal (III) weighing 3.2 g required hydrogen in a volume of 1.344 dm3 (n.c.). The metal was then dissolved in an excess of hydrochloric acid solution, and hydrogen was released in a volume of 0.896 dm3 (n.o.). Define the metal and write the equations for the corresponding reactions. 104 When reacting calcium halide weighing 0.200 g with a solution of silver nitrate, silver halide weighing 0.376 g was formed. Determine which calcium salt was used. 105 A mixture of sodium and potassium chlorides weighing 0.245 g was dissolved in water, and the resulting solution was treated with a solution of silver nitrate. As a result of the reaction, a precipitate with a mass of 0.570 g was formed. Calculate the mass fractions (%) of sodium and potassium chlorides in the mixture. 106 A mixture of sodium and lithium fluorides weighing 4 g was treated with concentrated sulfuric acid on heating. In this case, a mixture of metal sulfates weighing 8 g was obtained. Determine the content of salts in the initial mixture in mass fractions of a percent. 107 Determine the composition of the mixture (ω, %) of NaHCO3, Na2CO3, NaCl, if when it is heated with a mass of 10 g, a gas with a volume of 0.672 dm3 (n.o.) is released, and when interacting with hydrochloric acid of the same dm3 (n.o.s.). 108 Determine the composition of the mixture (ω, %) resulting from the interaction of powdered aluminum weighing 27 g with iron (III) oxide weighing 64 g. 109 After adding barium chloride to a solution containing a mixture of sodium and potassium sulfates weighing 1.00 g, barium sulfate was formed with a mass of 1.49 g. What is the ratio of sodium and potassium sulfates? 110 An excess of a solution of barium nitrate was added to an aqueous solution of aluminum and sodium sulfates weighing 9.68 g, and a precipitate weighed 18.64 g. Calculate the mass of aluminum and sodium sulfates in the initial mixture. 111 Interaction of an alloy of zinc and magnesium weighing 20 g with an excess solution of sulfuric acid formed a mixture of sulfates of these metals weighing 69 g. Determine the composition of the alloy in mass fractions of a percent. 112 An alloy of aluminum and magnesium weighing 3.00 g is mixed with an excess of chromium(III) oxide and ignited. As a result, chromium is formed with a mass of 5.55 g. Determine the composition of the initial mixture (ω, %). 113 A mixture of carbon monoxide and carbon dioxide with a volume of 1 dm3 (n.c.) has a mass of 1.43 g. Determine the composition of the mixture in volume fractions (%). 114 What mass of limestone containing calcium carbonate (ω = 90%) will be required to produce 10 tons of quicklime? 115 Treatment of a mixture of aluminum and aluminum oxide with a mass of 3.90 g with a NaOH solution gave off a gas with a volume of 840 cm3 (n.c.). Determine the composition of the mixture (ω, %). 1.4 Calculations according to the law of equivalents The amount of an element or substance that interacts with 1 mole of hydrogen atoms (1 g) or replaces this amount of hydrogen in chemical reactions is called the equivalent of this element or substance. Equivalent mass (Me) is the mass of 1 equivalent of a substance. Example 15 Determine the equivalent and equivalent masses of bromine, oxygen and nitrogen in the compounds HBr, H2O, NH3. Solution In these compounds, 1 mol of bromine atoms, 1/2 mol of oxygen atoms and 1/3 mol of nitrogen atoms combine with 1 mol of hydrogen atoms. Therefore, by definition, the equivalents of bromine, oxygen, and nitrogen are 1 mole, 1/2 mole, and 1/3 mole, respectively. Based on the molar masses of the atoms of these elements, we find that the equivalent mass of bromine is 79.9 g / mol, oxygen - 16 1/2 = 8 g / mol, nitrogen - 14 1/3 = 4.67 g / mol. The equivalent mass can be calculated from the composition of the compound if the molar masses (M) are known: 1) Me (element): Me = A / B, where A is the atomic mass of the element, B is the valency of the element; 2) Me (oxide) \u003d Me (elem.) + 8, where 8 is the equivalent mass of oxygen; 3) Me (hydroxide) \u003d M / n (he-), where n (he-) is the number of OH- groups; 4) Me (acids) \u003d M / n (n +), where n (n +) is the number of H + ions. 5) Me(salts) = M/nmeVme, where nme is the number of metal atoms; Vme is the valency of the metal. EXAMPLE 16 Determine the equivalent masses of the following substances Al, Fe2O3, Ca(OH)2, H2 SO4, CaCO3. Solution Me (Al) \u003d A / B \u003d 27/3 \u003d 9 g / mol; Me(Fe2O3) = 160/2 3 = = 26.7 g/mol; Me (Ca (OH) 2) \u003d 74/2 \u003d 37 g / mol; Me(H2SO4) = 98/2 = 49 g/mol; Me (CaCO3) \u003d 100/1 2 \u003d 50 g / mol; Me (Al2 (SO4) 3) \u003d 342/2 3 \u003d 342/6 \u003d 57 g / mol. Example 17 Calculate the equivalent mass of H2SO4 in the reactions: 1) H2SO4 + NaOH = NaHSO4 + H2O 2) H2SO4 + 2NaOH = Na2SO4 + H2O Solution The equivalent mass of a complex substance, as well as the equivalent mass of an element, can have different values, and depend on the chemical reaction that the substance enters into. The equivalent mass of sulfuric acid is equal to the molar mass divided by the number of hydrogen atoms substituted in this reaction for the metal. Therefore, Me(H2SO4) in reaction (1) is equal to 98 g/mol, and in reaction (2) - 98/2 = = 49 g/mol. When solving some problems containing information about the volumes of gaseous substances, it is advisable to use the value of the equivalent volume (Ve). Equivalent volume is the volume occupied under given conditions by 1 equivalent of a gaseous substance. So for hydrogen at n.o. the equivalent volume is 22.4 1/2 = 11.2 dm3, for oxygen - 5.6 dm3. According to the law of equivalents, the masses (volumes) of substances reacting with each other m1 and m2 are proportional to their equivalent masses (volumes): m1/ M e1 = m2/ M e2 . (1.4.1) If one of the substances is in the gaseous state, then: m/Me = V0/Ve. (1.4.2) Example 18 Combustion of a metal weighing 5.00 g produces metal oxide weighing 9.44 g. Determine the equivalent mass of the metal. Solution It follows from the conditions of the problem that the mass of oxygen is equal to the difference 9.44 g - 5.00 g = 4.44 g. The equivalent mass of oxygen is 8.0 g / mol. Substituting these values ​​into expression (1.4.1) we get: 5.00/Me(Me) = 4.44/8.0; M3(Me) = 5.00 8.0/4.44 = 9 g/mol. EXAMPLE 19 Oxidation of a metal(II) with a mass of 16.7 g produced an oxide with a mass of 21.5 g. Calculate the equivalent masses of: a) metal; b) its oxide. What is the molar mass of: c) metal; d) metal oxide? The solution m(O2) in the oxide will be: 21.54 - 16.74 = 4.80 g. In accordance with the law of equivalents, we get: 16.74 / Me (Me) = 4.80 / 8.00 Me) = 27.90 g/mol. The equivalent mass of the oxide is equal to the sum of the equivalent masses of the metal and oxygen and will be 27.90 + 8.00 = 35.90 g/mol. The molar mass of the metal (II) is equal to the product of the equivalent mass and the valence (2) and will be 27.90 2 = 55.80 g / mol. The molar mass of metal(II) oxide will be 55.8 + 16.0 = 71.8 g/mol. EXAMPLE 20 Metal nitrate weighing 7.27 g produces chloride weighing 5.22 g. Calculate the equivalent mass of the metal. Solution Since the equivalent mass of metal nitrate (chloride) is equal to the sum of the equivalent masses of metal (x) and the acid residue of nitrate (chloride), then according to the law of equivalents, taking into account the conditions of the problem, we obtain: 7.27 / 5.22 = (x + 62) / (x + 35.5). Where: x \u003d 32.0 g / mol. PRI mme R 21 From metal(II) sulfate weighing 15.20 g, a hydroxide weighing 9.00 g was obtained. Calculate the equivalent mass of the metal and determine the formula of the initial salt. Solution Taking into account the conditions of the problem and equation (1.4.1), we obtain: 15.2 / 9.0 = (Me (Me) + 48) / (Me (Me) + 17). Where: Me (Me) \u003d 28 g / mol; M (Me) \u003d 28 2 \u003d 56 g / mol. Salt formula: FeSO4. EXAMPLE 22 What mass of Ca(OH)2 contains as many equivalents as 312 g of Al(OH)3? The Me(Al(OH)3) solution is 1/3 of its molar mass, i.e. 78/3 = 26 g/mol. Therefore, 312 g of Al(OH)3 contains 312/26 = 12 equivalents. M e(Ca(OH)2) is 1/2 of its molar mass, i.e. 37 g/mol. Hence, 12 equivalents equals 37 12 = 444 g. EXAMPLE 23 The reduction of metal(II) oxide weighing 7.09 g requires hydrogen in a volume of 2.24 dm3 (N.O.). Calculate the equivalent masses of oxide and metal. What is the molar mass of the metal? Solution In accordance with the law of equivalents, we obtain: 7.09 / 2.24 = Me (oxide) / 11.20; Me(oxide) = 35.45 g/mol. The equivalent mass of the oxide is equal to the sum of the equivalent masses of the metal and oxygen, therefore, Me (Me) will be 35.45 - 8.00 \u003d 27.45 g / mol. The molar mass of metal (II) will be 27.45 2 = 54.90 g/mol. When determining the equivalent masses of various substances, for example, by the volume of gas released, the latter is collected over water. Then the partial pressure of the gas should be taken into account. The partial pressure of a gas in a mixture is the pressure that this gas would produce if it occupied the volume of the entire gas mixture under the same physical conditions. According to the law of partial pressures, the total pressure of a mixture of gases that do not enter into chemical interaction with each other is equal to the sum of the partial pressures of the gases that make up the mixture. If the gas is collected above the liquid, then in the calculations it should be borne in mind that its pressure is partial and is equal to the difference between the total pressure of the gas mixture and the partial pressure of the vapor of the liquid. EXAMPLE 24 What volume will be taken at n.o.s. 120 cm3 of nitrogen collected over water at 20°C and 100 kPa (750 mm Hg)? The saturation vapor pressure of water at 20 °C is 2.3 kPa. Solution The partial pressure of nitrogen is equal to the difference between the total pressure and the partial pressure of water vapor: PN 2 \u003d P - PH 2O \u003d 100 - 2.3 \u003d 97.7 kPa. Denoting the desired volume through V0 and using the combined Boyle-Mariotte and Gay-Lussac equation, we find: V0 = РVT0/TP0 = 97.7 120 273/293 101.3 = 108 cm3. Tasks 116 Calculate the equivalent and equivalent mass of phosphoric acid in the reactions of formation: a) hydrogen phosphate; b) dihydrophosphate; c) orthophosphate. 117 Determine the equivalent masses of sulfur, phosphorus and carbon in the compounds: H2S, P2O5, CO2. 118 An excess of potassium hydroxide was applied to solutions of: a) potassium dihydrogen phosphate; b) dihydroxovismuth(III) nitrate. Write the reaction equations for these substances with potassium hydroxide and determine their equivalents and equivalent masses. 119 Write the equations for the reactions of iron(III) hydroxide with hydrochloric (hydrochloric) acid, in which the following iron compounds are formed: a) dihydroxiron chloride; b) hydroxide iron dichloride; c) iron trichloride. Calculate the equivalent and equivalent mass of iron(III) hydroxide in each of these reactions. 120 Calculate the equivalent mass of sulfuric acid in the formation of: a) sulfate; b) hydrosulfate. 121 What is the equivalent volume (N.O.) of oxygen, hydrogen and chlorine? 122 Determine the equivalent mass of sulfuric acid if it is known that H2SO4 weighing 98 g reacts with magnesium weighing 24 g, whose equivalent mass is 12 g/mol. 123 During the combustion of magnesium weighing 4.8 g, oxide weighing 8.0 g was formed. Determine the equivalent mass of magnesium. 124 When a metal of mass 2.20 g reacted with hydrogen, a hydride of mass 2.52 g was formed. Determine the equivalent mass of the metal and write the formula for the hydride. 125 Determine the equivalent masses of tin in its oxides, the mass fraction of oxygen in which is 21.2% and 11.9%. 126 The reaction of a metal weighing 0.44 g required bromine weighing 3.91 g, the equivalent mass of which is 79.9 g/mol. Determine the equivalent mass of the metal. 127 The mass fraction of oxygen in lead oxide is 7.17%. Determine the equivalent mass of lead. 128 Mass fraction of calcium in chloride is 36.1%. Calculate the equivalent mass of calcium if the equivalent mass of chlorine is 35.5 g/mol. 129 Determine the equivalent mass of metal if the mass fraction of sulfur in sulfide is 22.15%, and the equivalent mass of sulfur is 16 g/mol. 130 The same mass of metal combines with oxygen with a mass of 0.4 g and with one of the halogens with a mass of 4.0 g. Determine the equivalent mass of the halogen. 131 Calculate the equivalent mass of aluminum if its combustion with a mass of 10.1 g produces oxide with a mass of 18.9 g. 132 The neutralization of oxalic acid (H2C2O4) with a mass of 1.206 g required KOH with a mass of 1.502 g, the equivalent mass of which is 56 g/mol. Calculate the equivalent mass of the acid. 133 To neutralize hydroxide weighing 3.08 g, hydrochloric acid weighing 3.04 g was consumed. Calculate the equivalent mass of hydroxide. 134 For the neutralization of phosphoric acid weighing 14.7 g, NaOH was consumed, weighing 12.0 g. Calculate the equivalent mass and basicity of phosphoric acid. Write an equation for the corresponding reaction. 135 To neutralize phosphorous acid (H3PO3) weighing 8.2 g, KOH weighing 11.2 g was consumed. Calculate the equivalent mass and basicity of phosphorous acid. Write the reaction equation. 136 To neutralize an acid weighing 2.45 g, NaOH weighing 2.00 g was consumed. Determine the equivalent mass of the acid. 137 A metal(I) oxide weighing 1.57 g contains a metal weighing 1.30 g. Calculate the equivalent mass of the metal and its oxide. 138 Calculate the atomic mass of metal(II) and determine what kind of metal it is if this metal weighing 8.34 g is oxidized by oxygen with a volume of 0.68 dm3 (n.o.). 139 Decomposition of a metal oxide weighing 0.464 g yielded a metal weighing 0.432 g. Determine the equivalent mass of the metal. 140 From a metal weighing 1.25 g, a nitrate of mass 5.22 g is obtained. Calculate the equivalent mass of this metal. 141 The interaction of aluminum weighing 0.32 g and zinc weighing 1.16 g with acid releases the same volume of hydrogen. Determine the equivalent mass of zinc if the equivalent mass of aluminum is 9 g/mol. 142 A metal chloride weighing 20.8 g produces sulfate of this metal weighing 23.3 g. Calculate the equivalent mass of the metal. 143 Metal nitrate weighing 2.62 g produces sulfate of this metal weighing 2.33 g. Calculate the equivalent mass of the metal. 144 From a metal iodide weighing 1.50 g, a nitrate of this metal weighing 0.85 g is obtained. Calculate the equivalent mass of the metal. 145 From metal sulfate weighing 1.71 g, hydroxide of this metal weighing 0.78 g is obtained. Calculate the equivalent mass of the metal. 146 From metal chloride weighing 1.36 g is obtained hydroxide of this metal weighing 0.99 g. Calculate the equivalent mass of the metal. 147 Metal nitrate with a mass of 1.70 g yields iodide of this metal with a mass of 2.35 g. Calculate the equivalent mass of the metal. 148 The interaction of a metal weighing 1.28 g with water produced hydrogen with a volume of 380 cm3 measured at 21°C and a pressure of 104.5 kPa (784 mm Hg). Calculate the equivalent mass of the metal. 149 What volume of hydrogen (n.c.) will be required to reduce a metal oxide weighing 112 g if the mass fraction of metal in the oxide is 71.43%? Determine the equivalent mass of the metal. 150 The equivalent mass of the metal is 23 g/mol. Determine the mass of metal that must be taken to extract hydrogen from an acid with a volume of 135.6 cm3 (n.o.s.). ). 151 Calculate the equivalent mass of the metal if a metal weighing 0.5 g displaces 184 cm3 of hydrogen from the acid, measured at 21 °C and a pressure of 101325 Pa. 152 Calculate the equivalent mass of the metal if a metal(II) with a mass of 1.37 g displaces 0.5 dm3 of hydrogen from the acid, measured at 18 °C and a pressure of 101325 Pa. 153 Determine the equivalent and atomic mass of the metal(II), if the reaction of a metal weighing 0.53 g with HCl yields H2 with a volume of 520 cm3 at 16 °C and a pressure of 748 mm Hg. Art. The pressure of saturated water vapor at a given temperature is 13.5 mm Hg. Art. 154 Metal(II) weighing 0.604 g displaced 581 cm3 of hydrogen from the acid, measured at 18 C and a pressure of 105.6 kPa and collected over water. The saturation vapor pressure of water at this temperature is 2.1 kPa. Calculate the atomic mass of the metal. 155 In a gasometer above water there is O2 with a volume of 7.4 dm3 at 296 K and a pressure of 104.1 kPa (781 mmHg). The pressure of saturated water vapor at this temperature is 2.8 kPa (21 mm Hg). What volume (n.c.) will the oxygen in the gasometer take up? 2 THE STRUCTURE OF THE ATOM AND THE PERIODIC SYSTEM OF DI MENDELEEV 2.1 The electron shell of the atom The motion of an electron in an atom is of a probabilistic nature. The circumnuclear space, in which an electron can be located with the highest probability (0.90 - 0.95), is called the atomic orbital (AO). An atomic orbital, like any geometric figure, is characterized by three parameters (coordinates) called quantum numbers (n, l , m l , ms). Quantum numbers do not take any, but certain, discrete (discontinuous) values. Neighboring values ​​of quantum numbers differ by one. Quantum numbers determine the size (n), shape (l), orientation (m l) of the atomic orbital in space. Atomic orbitals corresponding to l values ​​equal to 0, 1, 2, 3 are called s-, p-, d- and f-orbitals, respectively. In the electron-graphic formulas of atoms, each atomic orbital is denoted by a square (). Occupying one or another atomic orbital, an electron forms an electron cloud, which can have a different shape for electrons of the same atom. The electron cloud is characterized by four quantum numbers (n, l, m l, ms). These quantum numbers are related to the physical properties of the electron: the number n (principal quantum number) characterizes the energy (quantum) level of the electron; number l (orbital) - the moment of momentum (energy sublevel); number m l (magnetic) - magnetic moment; ms - spin. Spin arises due to the rotation of an electron around its own axis. According to the Pauli principle, an atom cannot have two electrons characterized by the same set of 4x quantum numbers. Therefore, in an atomic orbital there can be no more than two electrons that differ in their spins (ms = ± 1/2). In table. Table 1 shows the values ​​and designations of quantum numbers, as well as the number of electrons at the corresponding energy level and sublevel. The stable (unexcited) state of a many-electron atom corresponds to such a distribution of electrons over atomic orbitals, at which the energy of the atom is minimal. Therefore, they are filled in the order of successive increase in their energies. This filling order is determined by the Klechkovsky rule (rule n + l): - the filling of electronic sublevels with an increase in the atomic number of an element occurs from a lower value (n + l) to a higher value (n + l); - for equal values ​​(n + l), the energy sublevels with a smaller value of n are filled first. The sequence of filling energy levels and sublevels is as follows: 1s→ 2s→2p→ 3s→3p→ 4s→ 3d→ 4p→5s→ 4d→5p→ 6s→ (5d1) → → 4f →5d→ 6p →7s→ (6d1) →5f →6d→7p. The electronic structure of an atom can also be depicted in the form of schemes for distributing electrons in quantum (energy) cells, which are a schematic representation of atomic orbitals. The placement of electrons in atomic orbitals within one energy level is determined by the Hund (Hund) rule: electrons within an energy sublevel are first arranged one at a time, and then if there are more electrons than orbitals, then they are already filled with two electrons or so that the total spin is maximum. Example 25 Make electronic and electron-graphic formulas of atoms of elements with serial numbers 16 and 22. Solution Since the number of electrons in an atom of an element is equal to its serial number in the table of D. I. Mendeleev, then for sulfur -Z = 16, titanium - Z = 22. Electronic formulas are: 16S 1s 22s 22p 63s 23p4; 22Ti 1s 22s 22p 63s 23p 64s 23d 2. Electron graphic formulas of these atoms:

The human body contains about 5 g of iron, most of it (70%) is part of the hemoglobin in the blood.

Physical properties

In the free state, iron is a silvery-white metal with a grayish tint. Pure iron is ductile and has ferromagnetic properties. In practice, iron alloys are commonly used - cast irons and steels.

Fe is the most important and most common element of the nine d-metals of the secondary subgroup of group VIII. Together with cobalt and nickel, it forms the "iron family".

When forming compounds with other elements, it often uses 2 or 3 electrons (B \u003d II, III).

Iron, like almost all d-elements of group VIII, does not show a higher valency equal to the group number. Its maximum valency reaches VI and is extremely rare.

The most typical compounds are those in which the Fe atoms are in the +2 and +3 oxidation states.

Methods for obtaining iron

1. Commercial iron (in an alloy with carbon and other impurities) is obtained by carbothermal reduction of its natural compounds according to the scheme:

Recovery occurs gradually, in 3 stages:

1) 3Fe 2 O 3 + CO = 2Fe 3 O 4 + CO 2

2) Fe 3 O 4 + CO = 3FeO + CO 2

3) FeO + CO \u003d Fe + CO 2

The cast iron resulting from this process contains more than 2% carbon. In the future, steels are obtained from cast iron - iron alloys containing less than 1.5% carbon.

2. Very pure iron is obtained in one of the following ways:

a) decomposition of pentacarbonyl Fe

Fe(CO) 5 = Fe + 5CO

b) hydrogen reduction of pure FeO

FeO + H 2 \u003d Fe + H 2 O

c) electrolysis of aqueous solutions of Fe +2 salts

FeC 2 O 4 \u003d Fe + 2СO 2

iron(II) oxalate

Chemical properties

Fe - a metal of medium activity, exhibits general properties characteristic of metals.

A unique feature is the ability to "rust" in humid air:

In the absence of moisture with dry air, iron begins to noticeably react only at T > 150°C; when calcined, “iron scale” Fe 3 O 4 is formed:

3Fe + 2O 2 = Fe 3 O 4

Iron does not dissolve in water in the absence of oxygen. At very high temperatures, Fe reacts with water vapor, displacing hydrogen from water molecules:

3 Fe + 4H 2 O (g) \u003d 4H 2

The rusting process in its mechanism is electrochemical corrosion. The rust product is presented in a simplified form. In fact, a loose layer of a mixture of oxides and hydroxides of variable composition is formed. Unlike the Al 2 O 3 film, this layer does not protect the iron from further destruction.

Types of corrosion

Corrosion protection of iron

1. Interaction with halogens and sulfur at high temperature.

2Fe + 3Cl 2 = 2FeCl 3

2Fe + 3F 2 = 2FeF 3

Fe + I 2 \u003d FeI 2

Compounds are formed in which the ionic type of bond predominates.

2. Interaction with phosphorus, carbon, silicon (iron does not directly combine with N 2 and H 2, but dissolves them).

Fe + P = Fe x P y

Fe + C = Fe x C y

Fe + Si = FexSiy

Substances of variable composition are formed, since berthollides (the covalent nature of the bond prevails in the compounds)

3. Interaction with "non-oxidizing" acids (HCl, H 2 SO 4 dil.)

Fe 0 + 2H + → Fe 2+ + H 2

Since Fe is located in the activity series to the left of hydrogen (E ° Fe / Fe 2+ \u003d -0.44V), it is able to displace H 2 from ordinary acids.

Fe + 2HCl \u003d FeCl 2 + H 2

Fe + H 2 SO 4 \u003d FeSO 4 + H 2

4. Interaction with "oxidizing" acids (HNO 3 , H 2 SO 4 conc.)

Fe 0 - 3e - → Fe 3+

Concentrated HNO 3 and H 2 SO 4 "passivate" iron, so at ordinary temperatures the metal does not dissolve in them. With strong heating, slow dissolution occurs (without release of H 2).

In razb. HNO 3 iron dissolves, goes into solution in the form of Fe 3+ cations, and the acid anion is reduced to NO *:

Fe + 4HNO 3 \u003d Fe (NO 3) 3 + NO + 2H 2 O

It dissolves very well in a mixture of HCl and HNO 3

5. Attitude to alkalis

Fe does not dissolve in aqueous solutions of alkalis. It reacts with molten alkalis only at very high temperatures.

6. Interaction with salts of less active metals

Fe + CuSO 4 \u003d FeSO 4 + Cu

Fe 0 + Cu 2+ = Fe 2+ + Cu 0

7. Interaction with gaseous carbon monoxide (t = 200°C, P)

Fe (powder) + 5CO (g) \u003d Fe 0 (CO) 5 iron pentacarbonyl

Fe(III) compounds

Fe 2 O 3 - iron oxide (III).

Red-brown powder, n. R. in H 2 O. In nature - "red iron ore".

Ways to get:

1) decomposition of iron hydroxide (III)

2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O

2) pyrite roasting

4FeS 2 + 11O 2 \u003d 8SO 2 + 2Fe 2 O 3

3) decomposition of nitrate

Chemical properties

Fe 2 O 3 is a basic oxide with signs of amphoterism.

I. The main properties are manifested in the ability to react with acids:

Fe 2 O 3 + 6H + = 2Fe 3+ + ZN 2 O

Fe 2 O 3 + 6HCI \u003d 2FeCI 3 + 3H 2 O

Fe 2 O 3 + 6HNO 3 \u003d 2Fe (NO 3) 3 + 3H 2 O

II. Weak acid properties. Fe 2 O 3 does not dissolve in aqueous solutions of alkalis, but when fused with solid oxides, alkalis and carbonates, ferrites are formed:

Fe 2 O 3 + CaO \u003d Ca (FeO 2) 2

Fe 2 O 3 + 2NaOH \u003d 2NaFeO 2 + H 2 O

Fe 2 O 3 + MgCO 3 \u003d Mg (FeO 2) 2 + CO 2

III. Fe 2 O 3 - feedstock for iron production in metallurgy:

Fe 2 O 3 + ZS \u003d 2Fe + ZSO or Fe 2 O 3 + ZSO \u003d 2Fe + ZSO 2

Fe (OH) 3 - iron (III) hydroxide

Ways to get:

Obtained by the action of alkalis on soluble salts Fe 3+:

FeCl 3 + 3NaOH \u003d Fe (OH) 3 + 3NaCl

At the time of receipt of Fe(OH) 3 - red-brown mucosamorphous precipitate.

Fe (III) hydroxide is also formed during the oxidation of Fe and Fe (OH) 2 in humid air:

4Fe + 6H 2 O + 3O 2 \u003d 4Fe (OH) 3

4Fe(OH) 2 + 2Н 2 O + O 2 = 4Fe(OH) 3

Fe(III) hydroxide is the end product of hydrolysis of Fe 3+ salts.

Chemical properties

Fe(OH) 3 is a very weak base (much weaker than Fe(OH) 2). Shows noticeable acidic properties. Thus, Fe (OH) 3 has an amphoteric character:

1) reactions with acids proceed easily:

2) a fresh precipitate of Fe(OH) 3 is dissolved in hot conc. solutions of KOH or NaOH with the formation of hydroxo complexes:

Fe (OH) 3 + 3KOH \u003d K 3

In an alkaline solution, Fe (OH) 3 can be oxidized to ferrates (salts of iron acid H 2 FeO 4 not isolated in the free state):

2Fe(OH) 3 + 10KOH + 3Br 2 = 2K 2 FeO 4 + 6KBr + 8H 2 O

Fe 3+ salts

The most practically important are: Fe 2 (SO 4) 3, FeCl 3, Fe (NO 3) 3, Fe (SCN) 3, K 3 4 - yellow blood salt \u003d Fe 4 3 Prussian blue (dark blue precipitate)

b) Fe 3+ + 3SCN - \u003d Fe (SCN) 3 Fe (III) thiocyanate (blood red solution)

through x and y. CrO oxide formula. The oxygen content in chromium oxide is 31.6%. Then:

x: y \u003d 68.4 / 52: 31.6 / 16 \u003d 1.32: 1.98.

To express the resulting ratio as integers, we divide the resulting numbers by a smaller number:

x: y \u003d 1.32 / 1.32: 1.98 / 1.32 \u003d 1: 1.5,

and then multiply both values ​​of the last ratio by two:

Thus, the simplest formula of chromium oxide is Cr2O3.

EXAMPLE 10 With the complete combustion of a certain substance weighing 2.66 g, CO2 and SO2 were formed with masses of 1.54 g and 4.48 g, respectively. Find the simplest formula of a substance.

Solution The composition of the combustion products shows that the substance contained carbon and sulfur. In addition to these two elements, oxygen could also be included in its composition.

The mass of carbon that was part of the substance can be found from the mass of CO2 formed. The molar mass of CO2 is 44 g/mol, while 1 mole of CO2 contains 12 g of carbon. Find the mass of carbon m contained in 1.54 g of CO2:

44/12 = 1.54/m; t \u003d 12-1.54 / 44 \u003d 0.42 g.

Calculating in a similar way the mass of sulfur contained in 4.48 g of SO2, we obtain 2.24 g.

Since the mass of sulfur and carbon is 2.66 g, this substance does not contain oxygen and the formula of the substance is Cx8y:

x: y \u003d 0.42 / 12: 2.24 / 32 \u003d 0.035: 0.070 \u003d 1: 2.

Therefore, the simplest formula of the substance is C82. To find the molecular formula of a substance, it is necessary, in addition to the composition of the substance, to know its molecular weight.

PRI me R 11 Gaseous connection of nitrogen with hydrogen contains 12.5% ​​(wt.) hydrogen. The density of the compound by hydrogen is 16. Find the molecular formula of the compound.

Solution The desired formula of the substance NJH,:

x: y \u003d 87.5 / 14: 12.5 / 1 \u003d 6.25: 12.5 \u003d 1: 2.

The simplest formula of the compound NH2. This formula corresponds to a molecular weight equal to 16 a. e. m. We find the true molecular weight of the compound, based on its hydrogen density:

M = 2-16 = 32 amu

Therefore, the formula of the substance is ^H4.

PRI mme R 12 When calcining hydrated zinc sulfate weighing 2.87 g, its mass decreased by 1.26 g. Set the formula of crystalline hydrate.

Solution When calcined, the crystalline hydrate decomposes:

ZnSO4 - pH2O ZnSO4 + nH2Ot ; М(ZnSO4) = 161 g/mol; M(H2O) = 18 g/mol.

It follows from the condition of the problem that the mass of water is 1.26 g, and the mass of ZnSO4 is (2.87 - 1.26) = 1.61 g. Then the amount of ZnSO4 will be: 1.61/161 = 0.01 mol, and the number moles of water 1.26/18 = 0.07 mol.

Therefore, there are 7 moles of H2O per 1 mol of ZnSO4 and the formula of crystalline hydrate is ZnSO4-7H2O.

PRI mme R 13 In a stream of chlorine burned a mixture of copper and iron filings weighing 1.76 g; resulting in a mixture of metal chlorides weighing 4.60 g. Calculate the mass of copper that reacted.

Solution The reactions proceed according to the schemes:

1) Cu + CI2 = CuCl2;

2) 2Fe + 3Cl2 = 2FeCl3; M(Ou) = 64 g/mol; M^) = 56 g/mol;

M^uOD = 135 g/mol; M^a^ = 162.5 g/mol.

According to the condition of the problem, the mass of the mixture of copper(II) and iron(III) chlorides, i.e. a + b \u003d 4.60 g. Hence 135x / 64 + 162.5 - (1.76 - x) / 56 \u003d 4.60.

Therefore, x \u003d 0.63, that is, the mass of copper is 0.63 g.

EXAMPLE 14 When a mixture of potassium hydroxide and hydrogen carbonate was treated with an excess of hydrochloric acid solution, potassium chloride was formed weighing 22.35 g and a gas with a volume of 4.48 dm3 (n.o.) was released. Calculate the composition (co, %) of the initial mixture.

Solution of the Reaction Equation

1) KHCO3 + HCl = KCl + H2O + CO2t;

2) KOH + HCl = KCl + H2O;

M(KHCO3) = 100 g/mol; M(t) = 74.5 g/mol; M(WSH) = 56 g/mol.

According to the condition of the problem, the volume of gas (CO2) according to reaction (1) is 4.48 dm3 or 0.2 mol. Then from the reaction equation (1) it follows that the initial amount in a mixture of potassium bicarbonate is 0.2 mol or 0.2-100 = 20 g and the same amount is formed 0.2 mol KCl or 0.2-74.5 = 14 .9

Knowing the total mass of KO formed as a result of reactions (1) and (2), it is possible to determine the mass of KS1 formed by reaction (2). It will be 22.35 - 14.9 \u003d 7.45 g or 7.45 / 74.5 \u003d 0.1 mol. The formation of 0.1 mol of KS1 according to reaction (2) will require the same amount of KOH, that is, 0.1 mol or 0.1 -56 = 5.60 g. Therefore, the content of the initial components in the mixture will be:

5.6-100 / 25.6 = 21.9% KOH and 20.0-100 / 25.6 = 78.1% KHCO3.

EXAMPLE 15 When calcining barium iodide weighing 4.27 g, a precipitate weighing 3.91 g remained. Determine the mass fraction of barium iodide in the solution obtained by dissolving this crystalline hydrate weighing 60 g in water with a volume of 600 cm3.

Solution Equation of crystalline hydrate dehydration

BaJ2 - xH2O = BaJ2 + x^Ot . The molar mass of BaJ2 is 391 g/mol, and the molar mass of crystalline hydrate BaJ2 - xH2O - (391 + 18x) g/mol. Let's make a proportion:

(391 + 18x) g crystalline hydrate - 391 g anhydrous salt

4.27 g crystalline hydrate - 3.91 g anhydrous salt

we find that x \u003d 2. Thus, the formula of BaJ2 crystalline hydrate is 2H2O. Its molar mass is 427 g/mol. A crystalline hydrate weighing 60 g contains 60-391/427 = 54.9 g of anhydrous salt. Let us calculate the mass fraction of barium iodide in a solution obtained by dissolving BaJ2 - 2H2O weighing 60 g in water with a volume of 600 cm3:

ro(BaJ2) = m(BaJ2) / = 54.9 / (600 + 60) = 0.083.

EXAMPLE 16 When a metal (11) with a mass of 2.18 g is oxidized with oxygen, a metal oxide with a mass of 2.71 g is obtained. What is this metal?

Solution Metal(11) oxide has an EO composition and its molar mass is equal to the sum of the atomic masses of the metal and oxygen. Let the atomic mass of the metal be x g/mol, then the molar mass of the metal oxide is (x + 16) g/mol. Considering the conditions of the problem, we compose the proportion:

x g of metal - (x + 16) g of metal oxide

2.18 g metal - 2.71 g metal oxide

whence x \u003d 65.8 g. Therefore, the metal is zinc.

Example 17 Determine the true formula of a gaseous substance that contains fluorine and oxygen with mass fractions of 54.29% and 45.71%, respectively, if its relative density in nitrogen is

Solution The test substance weighing 100 g contains fluorine and oxygen with masses of 54.29 g and 45.71 g, respectively. Find the number of moles of atoms:

54.29/19 = 2.86 moles of fluorine and 45.71/16 = 2.86 moles of oxygen.

Thus, the number of fluorine atoms in a molecule is equal to the number of oxygen atoms. Therefore, the simplest formula of a substance is (OF)n. This formula corresponds to a molar mass equal to 35 g/mol. We find the true molar mass of a substance based on its nitrogen density: M = 28-2.5 = 70 g / mol. Let's write the equation for the molar mass of the compound (OF)n: 16n + 19n = 70. Whence n = 2.

Therefore, the formula of the substance is O2F2.

PRI mme R 18 Set the formula of a mineral having a composition in mass fractions of a percent: silicon - 31.3; oxygen - 53.6; a mixture of aluminum and beryllium - 15.1.

Solution Electroneutrality Equation:

(+4) - 31.3 / 28 + (-2) - 53.6 / 6 + (+3) - x / 27 + (+2) - (15.1 - x) / 9 = 0 4.47 - 6.7 + 0.11x + 3.356 - 0.222x = 0 x = 10.14.

Therefore, the mineral contains aluminum - 10.14%; beryllium - 4.96

Find the number of silicon, oxygen, beryllium and aluminum atoms:

31,3/28: 53,6/16: 10,14/27: 4,96/9 = 1,118: 3,350: 0,375: 0,551.

To express the resulting ratios as integers, we divide the resulting numbers by a smaller number:

Therefore, the mineral formula is Si6Al2Be3O18 or Al2O3-3BeO-6SiO2.

EXAMPLE 19 A mixture of hydrogen and an unknown gas with a volume of 10 dm3 (n.a.) has a mass of 7.82 g. Determine the molar mass of an unknown gas, if it is known that to obtain all the hydrogen in the mixture, metal zinc weighing 11.68 g in its reaction with sulfuric acid.

Solution Reaction Equation:

Zn + H2SO4 = ZnSO4 + H2t The number of moles of hydrogen is equal to the number of moles of zinc

n (H2) \u003d n (Zn) \u003d 11.68 / 65 \u003d 0.18 mol.

The volume (n.o.s.) and mass of hydrogen, respectively, are:

V (H2) = 0.18-22.4 = 4.03 dm3; m (H2) \u003d 0.18-2 \u003d 0.36 g. Find the volume, mass and molar mass of an unknown gas: V (gas) \u003d 10 - 4.03 \u003d 5.97 dm3; m (gas) \u003d 7.82 - 0.36 \u003d 7.46 g; M (gas) \u003d 7.46-22.4 / 5.97 \u003d 28 g / mol.

PRI me R 20 When burning organic matter weighing 7.2 g, the vapor density of which is equal to 36 for hydrogen, carbon monoxide (IV) and water were formed with masses of 22 g and 10.8 g, respectively. Determine the formula of the starting material.

Solution The equation for the combustion of organic matter of unknown composition:

CjyA, + O2 = xCO2 + y/2 H2O + zA

M(CO2) = 44 g/mol; M(H2O) = 18 g/mol.

Let's find the masses of hydrogen and carbon in the substance: m (H2) = n (H2) - M(H2) = m (H2O) - M(H2) / M(H2O) = 10.8-2 / 18 = 1.2 g ; m (C) \u003d n (C) - A (C) \u003d m (CO2) - A (C) / M (CO2) \u003d 22-12 / 44 \u003d 6.0 g.

Since the total mass of carbon and hydrogen is equal to the mass of the burned substance, a hydrocarbon of the composition CxHy was burned. We find the true molecular weight of a hydrocarbon based on its hydrogen density: M = 2-36 = 72 a.m.u.

To establish the formula of a hydrocarbon, we compose a proportion:

7.2 g CxHy - 22 g CO2

72 g CxHy - 44 g CO2.

Hence, x = 5, that is, the CxHy molecule contains 5 carbon atoms. The number of hydrogen atoms is (72 - 12-5) / 1 = 12. Therefore, the formula of organic matter is C5H12.

61 During the decomposition of metal(11) carbonate with a mass of 21.0 g, СО2 was released with a volume of 5.6 dm3 (n.o.). Set the salt formula.

62 Find the formulas of compounds having a composition in mass fractions of a percent:

a) sulfur - 40 and oxygen - 60;

b) iron - 70 and oxygen - 30;

c) chromium - 68.4 and oxygen - 31.6;

d) potassium - 44.9; sulfur - 18.4 and oxygen - 36.7;

e) hydrogen - 13.05; oxygen - 34.78 and carbon - 52.17;

f) magnesium - 21.83; phosphorus - 27.85 and oxygen - 50.32.

63 Determine the formulas of compounds that have a composition in mass

fractions of a percent:

a) potassium - 26.53; chromium - 35.35 and oxygen - 38.12;

b) zinc - 47.8 and chlorine - 52.2;

c) silver - 63.53; nitrogen - 8.24 and oxygen - 28.23;

d) carbon - 93.7; hydrogen - 6.3.

64 Determine the simplest formulas of minerals that have a composition in

mass fractions of a percent:

a) copper - 34.6; iron - 30.4; sulfur - 35,

To establish the truth of the formula in the problem, there may be additional information of a different nature. Most often, this is the molar mass of the substance (in explicit or implicit form) and / or a description of the transformations into which it enters.

Example: 12

During the combustion of 0.88 g of some organic compound, 0.896 l of carbon dioxide and 0.72 g of water were formed. The vapor density of a given compound with respect to hydrogen is 44. What is the true formula of an organic compound?

First of all, let us clarify what elements the molecule of the substance under study consists of. Since the combustion products contain and , we conclude that the burned substance contains carbon and hydrogen. Let's find their number.

22.4 l is formed from 12 g

18 g is formed from 2 g

Find out if the substance under study contains oxygen.

Add the found masses of hydrogen and carbon
0.48 + 0.08 \u003d 0.56 g and is comparable with the initial mass of the burnt substance \u003d 0.88 g. Since , then the substance contains oxygen

Let us find the simplest formula in the form

The simplest formula.

We will look for the true formula in the form .

We find the true molar mass using the relative density of the organic compound with respect to hydrogen. The relative density of one gas (A) over another (B) is found by the formula.

hence, ,

where 2 is the molar mass of hydrogen, g/mol;

Relative density of an organic compound with respect to hydrogen.

44 2=88 g/mol

n= 88/44=2, therefore, the true formula of the substance is:

(C 2 H 4 O) 2 or C 4 H 8 O 2

Example 13

Task A. I. Zhirov (9-3) IV stage of the GOS in Chemistry 2004

The table shows the compositions of four binary compounds having the same qualitative composition.

Composition Compound
I	93,10	6,90
II	87,08	12,92
III	83,49	16,51
IV	81,80	18,20

1. Determine the qualitative composition of compounds (A, B).

2. Determine the composition of compounds I–IV (formulas).

1. Analysis of the condition of the problem

The condition says that all compounds consist of two identical elements. Since , it can be assumed that, therefore, A is most likely a non-metal located in the 2nd period, or hydrogen, therefore their formulas can be represented as:

Thus, A is necessarily a non-metal, and B can be either a metal or a non-metal.

2. Finding the coefficients x, y, z, w

To find them, we use the law of multiple ratios: “If two elements can form several compounds among themselves, then the mass fractions of any of the elements in these compounds, related to the mass fraction of the other, are treated as small integers.”

We divide by the largest number 13.49, since in real compounds the amount of B is not the same, but increases from the first compound to the fourth.

.

3. Determination of the atomic mass of element B

We will determine element B by enumeration of possible non-metals A. There are not many of them. These are H, F, having a valence of 1, and O, having a valence of 2. The remaining non-metals of the second period cannot form more than two compounds with one element.

Consider monovalent non-metals, for them the formulas of compounds will take the form:

II V 0.5 A or VA 2

III B 3/8 A or B 3 A 8

IV B 1/3 A or BA 3.

Calculate the atomic mass of element B using the data for the 1st compound.

If A - "N", then

A B \u003d 13.49 A H \u003d 13.49 1 \u003d 13.49, since

.

There is no such element.

If A is F, then

And B \u003d 13.49 19 \u003d 256.31 - there is no such element.

Consider a divalent non-metal - oxygen. For him, the formulas of the compounds will take the form.

IV B 2 A 3, since for divalent oxygen the compounds will have the form B 2 A n, it follows that B 2 A n \u003d B 2 1 / n A.

Consider the atomic mass of element B, based on the data for the II compound

And B \u003d 6.74 16 \u003d 107.84 is Ag (silver).

Answer I Ag 2 O

Example 14

Task (9-1) A. I. Zhirov 2004 IV stage of the GOS in Chemistry

The mineral moissanite was named after the French chemist Henri Moissan. Moissanite is highly chemically resistant to most chemicals and has a high refractive index. In modern jewelry, cut moissanite inserts replace diamonds. “... But by fusion with caustic alkalis in a silver crucible, transfer to a solution is easily removed ...” (F. Tredwell, “Course of Analytical Chemistry”, vol. 1, p. 319, Odessa, 1904.)

A portion of 1.000 g of finely ground moissanite was fused in a silver crucible with 7.0 g of sodium hydroxide monohydrate. The resulting melt was completely dissolved in 50 ml of water. With the careful addition of 30 ml of a 20% hydrochloric acid solution (density 1.1 g / cm 3) to the resulting solution, 0.56 l (n.o.) of gas was released, the density of which in air is 1.52, and white precipitated sediment. The precipitate was separated by filtration, washed with distilled water, and calcined at 900°C. Its weight after calcination was 1.500 g. The entire filtrate was evaporated to dryness, the weight of the dry residue was 7.05 g.

1. determine the composition of moissanite (formula).

2. Write the reaction equation for the conversion of moissanite into a soluble state (fusion with alkalis). What gaseous products can be released in this reaction?

3. Why do you think it is more convenient to use sodium hydroxide monohydrate for fusion?

4. Write the equations for the reactions that occur when an acid is added to the analyzed solution.

5. Write the reaction equations for obtaining a synthetic analogue of moissanite in the laboratory.

1. Analysis of the condition of the problem

Since the moissanite mineral does not dissolve under normal conditions either in acids or alkalis, and its transfer into solution is possible only after fusion with alkali, it can be assumed that it is formed either only from non-metals or from an amphoteric metal that forms a cation. Imagine the chemical formula of the mineral A x B y. We represent the analysis scheme in the form

A x B y + NaOH H 2 O → PLAV + H 2 O → Solution without sediment

and outgassing + HCl → gas precipitate + solution calcined precipitate + dry residue

2. Calculate the molar mass and the amount of released gas.

D air \u003d 1.52 M gas \u003d 1.52 29 \u003d 44.08 g / mol

.

Gases having a molar mass of 44 g / mol are CO 2, N 2 O, C 3 H 8, CH 3 COH. Of these gases, CO 2 is most suitable, since only CO 2 can be isolated from alkaline melts by the action of HCl.

3. Determine the substance in excess in the reaction of the melt with a solution of HCl.

3.1. Calculate the number of moles of sodium hydroxide monohydrate

.

3.2. Calculate the number of moles of added acid

.

Assume that the dry residue consists only of NaCl, then

,

as you can see = , it follows that the entire dry residue consists only of NaCl, therefore, the elements that formed the mineral passed into CO 2 gas and sediment.

When a mineral is fused with alkali, the latter is always taken in excess, otherwise the alloy will not completely dissolve in water, so the elements that formed the mineral passed into the anions of sodium salts. For the formation of 0.12 moles of NaCl, 0.12 moles of HCl was sufficient, therefore, HCl was taken in excess, so all the carbon contained in the mineral was released as CO 2 .

4. determination of the qualitative and quantitative composition of the mineral.

There can be no CO 3 2- anion in the mineral, since carbonates dissolve well in acids, so we will calculate by carbon.

Suppose that there is one carbon atom in the anion of the mineral, then 1 g of the mineral contains 0.025 mol of the compound

.

As we see,< , что еще раз подтверждает, что минерал не относится к классу карбонатов. При такой маленькой молярной массе, анионом может быть только С 4- , тогда х·М А = 40–12 = 28. Если х = 1, тогда А – Si, если х = 2,
A - N. Nitrogen disappears, since there is no CN 2 compound, so most likely A is Si, and moissanite has the formula SiC.

Let's check this assumption by obtaining the sediment formula

is SiO 2 .

Indeed, according to the reactions, SiO 2 is obtained:

SiC + 4 NaOH H 2 O \u003d Na 2 SiO 3 + Na 2 CO 3 + 2H 2 O + 4H 2

H 2 O + Na 2 SiO 3 + 2HCl → 2NaCl + H 4 SiO 4

(n-2)H 2 O + H 4 SiO 4 → SiO 2 nH 2 O↓

SiO 2 nH 2 O SiO 2 + H 2 O

Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O.

7 Standard multivariate tasks

1 Determine the molar mass of the metal equivalent, ... grams of which are displaced from acid ... hydrogen, collected over water at a temperature ... °C and adding ... (Saturated water vapor pressure see appendix). Numerical data for all options are offered in Table. one.

2 Oxidation of ... grams of A (...) spent ... ml (l) oxygen measured under normal conditions. Determine the molar mass of the equivalent of the element (A), the percentage composition of the resulting oxide, its formula, and use chemical reactions to show its acid-base character. Numerical data for all options are offered in Table. 2.

3 Determine mass 1 m 3 gas mixture of the specified composition under the following conditions. Numerical data for all options are offered in Table. 3.

5 Determine the masses and volumes (for gaseous substances) after the completion of the reaction (...) substance A with (...) substance B (...). Numerical data for all options are proposed in Table. 5.

6 A mixture of (...) grams, consisting of substance A and B (...), was treated with an excess of hydrochloric acid. At the same time, (…) liters of hydrogen, measured at n.c., were released. Determine the mass fraction of each component of the mixture. Numerical data for all options are offered in Table. 6.

7 Burning (…) produced (…) water and (…) carbon dioxide (n.c.). Find the true formula of an organic compound if the relative density of its vapor is (...). Numerical data for all options are offered in Table. 7.

Table 1

Numerical data for problem 1

Option	Weight of metal, g	Volume of hydrogen, l	Temperature, °C	Pressure
	5,4000	7,70000	27,0	756.7 mmHg
	0,5840	0,21900	17,0	754.5 mmHg
	0,5000	0,18450	21,0	1.0 atm.
	0,1830	0,18270	20,0	767.5 mmHg
	1,1500	0,62300	20,0	751.5 mmHg
	0,0600	0,06050	14,0	752 mmHg
	2,7900	0,62300	20,0	751.5 mmHg
	11,1700	7,70000	27,0	756.5 mmHg
	0,6500	0,25400	29,0	1.0 atm.
	0,2700	0,38500	27,0	756.7 mmHg
	0,5870	0,25400	29,0	1.0 atm.
	0,1200	0,12100	14,0	0.99 atm.
	0,4600	0,24900	20,0	0.99 atm.
	0,2500	0,09225	21,0	1.0 atm.
	0,2046	0,27400	19,0	771.5 mmHg
	1,1100	0,40420	19,0	770 mmHg
	0,3470	0,18000	15,0	0.85 atm.
	0,0750	0,02850	22,0	745 mmHg
	0,0230	0,03230	19,5	763 mmHg
	0,1110	0,04042	19,0	1.01 atm.
	0,5400	0,77000	27,0	0.96 atm.
	0,2500	0,09225	21,0	760 mmHg
	0,2400	0,24200	14,0	753 mmHg
	1,0000	0,36900	21,0	760 mmHg
	0,6000	0,06050	14,0	0.9 atm.

table 2

Numerical data for problem 2

Option	Mass of substance A, g	Oxygen volume
	1.24 phosphorus	672 ml
	1.27 copper	0.224 l
	0.92 sodium	224 ml
	0.5 sulfur	0.35 l
	1.92 molybdenum	672 ml
	15.6 potassium	2.24 l
	2.43 magnesium	1.12 l
	2.24 iron	672 ml
	2.0 carbon	3.78 l
	0.41 phosphorus	0.373 l
	3.95 selenium	1.68 l
	9.2 lithium	7.46 l
	0.8 calcium	224 ml
	2.8 gallium	0.672 l
	1.0 sulfur	700 ml
	0.954 copper	0.168 l
	0.766 tungsten	140 ml
	2.28 germany	0.7 l
	4.88 antimony	0.672 l
	0.6 arsenic	0.224 l
	0.14 nitrogen	112 ml
	2.08 chrome	672 ml
	0.544 niobium	0.224 l
	3.63 germany	1.12 l
	0.724 tantalum	112 ml

Table 3

Numerical data for problem 3

Option	Gas	%	Gas	%	Gas	%	T, °С	Pressure
	About 2		N 2		Ar			1 atm.
	About 2		N 2		CO2			730 mm Hg
	SO		CO2		N 2			100 kPa
	SO		H2O		CO2			1.4 atm.
	SO		O2		N 2			740 mm Hg
	C 2 H 4		CH 4		CO			700 mm Hg
	Cl 2		H2		HCl			800 mm Hg
	O2		Ar		N 2			110 kPa
	O2		H2		Ar			98 kPa
	O2		H2		Ne		-20	1.1 atm.
	Cl2		N 2		H2			10 kPa
	N 2		CH 4		C 2 H 2			760 mm Hg
	F2		O2		-	-	-100	0.1 atm.
	He		Ar		O2			780 mm Hg
	kr		N 2		Cl2			500 mm Hg
	H2		O2		-	-		90 kPa
	Br2		N 2		Ar			1 MPa
	CO		CO2		N 2			140 kPa
	CO		CO2		O2			0.8 atm.
	CH 4		CO		H2			0.9 atm.
	C 2 H 4		CH 4		C 2 H 6			745 mmHg
	H2		He		O2			1.9 atm.
	CO		H2		H2O			1.0 atm.
	Cl2		TiCl 4		N 2			730 mm Hg
	CO		Fe(CO)5		-	-		100 atm.

Table 4

Numerical data for problem 4

Option	Concentrate	Substance B	reaction product
Mineral A	Weight, g	Substance D	Volume, l
	Bi 2 S 3		O2	SO2	22,4
	PbS		O2	SO2	2,24
	Ag2S		O2	SO2	11,2
	FeS		O2	SO2	15,68
	FeS 2		O2	SO2	5,6
	NiS		O2	SO2	1,12
	Cu 2 S		O2	SO2	2,24
	ZnS		O2	SO2	44,8
	HgS		O2	SO2	33,6
	M&S 2		O2	SO2	3,36
	PTS 2		O2	SO2	1,792
	Fe2O3		C	CO(Fe)	13,44
	Fe 3 O 4		C	CO(Fe)	8,96
	FeO		C	CO(Fe)	11,2
	Fe2O3		H2	H 2 O (Fe)	6,72
	C		O2	CO2	33,6
	ZnO		C	CO	22,4
	WS 2		O2	SO2	7,2
	MoS 2		O2	SO2	11,2
	GeS		O2	SO2	1,12
	VS 2		O2	SO2	5,6
	Cu2O		C	CO	1,12
	Ag2O		C	CO	2,24
	FeS 2		O2	SO2	17,92
	Cu 2 S		O2	SO2	1,586

Table 5

Numerical data for problem 5

Option	Substance A	Weight, g	Substance B	Mass or volume	Pressure	Temperature, °C
	Fe		HCl	50 g 10%	740 mm
	Na2CO3		CaCl2	100 g 40%	-	-
	NaOH		H2SO4	50 g	-	-
	CaCO3		HCl	100 g 5%	800 mm
	NaOH		CO2	14 l	1.2 atm
	NaCl		AgNO3		-	-
	FeCl3		NaOH	200 g 20%	-	-
	CuSO4		H 2 S	10 l	100 kPa
	Pb(NO 3) 2		H 2 S	20 l	1 MPa
	MgCO3		HNO3	100 g 60%	10 kPa
	KOH		HCl	300 g 20%	-	-
	KOH		HCl	10 l	730 mm
	BaCl2		H2SO4	500 g 25%	-	-
	BaCl2		AgNO3	600 g	-	-
	FeCl3		AgNO3	1000 g	-	-
	Ba(OH)2		HCl	5 l	1.0 atm
	LiOH		CO2	30 l	0.1 atm
	Cu(NO 3) 2		NaOH	1 kg 10%	-	-
	Al 2 (SO 4) 3		NH4OH	5 kg 10%	-	-
	CaCl2		Na2CO3	1 kg 20%	-	-
	Ba(OH)2		Na3PO4	5 kg 2%	-	-
	Hg(NO 3) 2		H 2 S	50 l	1.2 atm
	CdCl2		H 2 S	80 l	700 mm
	K2CO3	1 000	HCl	100 l	101 kPa
	Na2CO3		H2SO4	5 kg 30%	110 kPa

Table 6

Numerical data for problem 6

Option	Mass of the mixture, g	Substance A	Substance B	Hydrogen volume, l
	8,5	Na	K	3,235
	4,71	La(La + 3)	Al	1,68
	7,27	Zn	ZnO	1,25
	8,0	Fe	mg	4,48
	12,0	Al	Al2O3	3,73
	3,32	La(La + 3)	Al	1,34
	4,445	mg	Al	4,77
	4,5	Ti(Ti+3)	TiO2	0,46
	1,31	mg	Al	1,3
	1,5	Cu	mg	0,56
	1,32	Zn	SiO2	0,44
	2,5	Zn	mg	1,4
	31,045	K	Na	11,2
	18,659	Fe	Zn	6,72
	37,46	Ca	Al	24,64
	10,0	Fe	FeO	2,24
	14,262	Li	Al	19,040
	6,755	Ba	Ca	2,688
	19,99	Ca	CaCl2	10,060
	11,933	Zn	Al	8,96
	6,484	sc	Fe	2,912
	176,442	Ba	K	33,6
	38,324	Cr(Cr+3)	Fe	17,92
	20,24	Cr(Cr+3)	Cr2O3	6,72
	252,5	Ba	BaO	22,4

Table 7 = 21

2.2 g org. conn. 1.8 ml 2.24 l = 2,75

8 Advanced tasks

Tasks on the mixture

1 A mixture of calcium carbide and calcium carbonate was treated with an excess of hydrochloric acid, resulting in a mixture of gases with an air density of 1.27 and a solution, upon evaporation of which a solid residue of mass 55.5 was formed G. Determine the mass of the initial mixture and the mass fractions of substances in it.

2 A mixture of calcium carbide and aluminum carbide was treated with excess water, resulting in a mixture of gases with an ammonia density of 1.0. After evaporation of the resulting solution, a precipitate was obtained, upon calcination of which a solid residue was formed with a mass of 66.8 G. Determine the mass of the initial mixture and the mass fractions of the substances in it.

3 When calcining a mixture of mass 41 G, consisting of sodium acetate and an excess of sodium hydroxide, a gas was released that reacted with chlorine when illuminated. As a result of the last reaction, 11.95 G trichloromethane (chloroform). The yield of chloroform was 40 % from the theoretical. Find the mass fractions of substances in the original mixture.

4 When calcining a mixture of iron (II) nitrates and mercury, a gas mixture was formed, which by 10 %

5 When calcining a mixture of iron (II) and iron (III) nitrates, a gas mixture was formed, which by 9 % heavier than argon. How many times did the mass of the solid mixture decrease after calcination?

6 For dissolution 1.26 G magnesium aluminum alloy used 35 ml sulfuric acid solution (mass fraction 19.6 % , density 1.14). The excess acid reacted with 28.6 ml potassium bicarbonate solution with a concentration of 1.4 mol/l. Calculate the mass fractions of metals in the alloy and the volume of gas (at N.C.) released during the dissolution of the alloy.

7 Silver-copper melt sample, weight 3.54 G, density dissolved in 23.9 ml nitric acid solution (mass fraction of acid 31.5 % , solution density 1.17). It took 14.3 ml barium hydroxide solution with a concentration of 1.4 mol/l. Calculate the mass fractions of metals in the alloy and the volume of gas (at N.C.) released during the dissolution of the alloy.

8 A mixture of iron and zinc filings, weighing 2.51 G, processed 30.7 ml sulfuric acid solution (mass fraction of acid 19.6 % , solution density 1.14). It took 25 ml potassium bicarbonate solution with a concentration of 2.4 mol/l. Calculate the mass fractions of metals in the initial mixture and the volume of gas (at N.C.) released during the dissolution of metals.

9 A mixture of barium sulfate and carbon, weighing 30 G, calcined without access to oxygen at a temperature of 1200 °C. The product obtained after calcination was treated with an excess of hydrochloric acid. The mass of the undissolved precipitate was 1.9 G. Write down the equations of the corresponding reactions and determine the mass fractions of substances in the initial mixture.

10 Calculate the mass fractions of the components of a mixture consisting of ammonium hydrogen carbonate, calcium carbonate and ammonium hydrogen phosphate, if it is known that from 62.2 G this mixture received 17.6 G carbon monoxide (IV) and 10.2 G gaseous ammonia.