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22 Chemistry of the Nonmetals Solutions to Exercises (c) 1 X 9.1 10⁶ g g O₂ H₂O 1 1.0 g H₂O X 1000 1L mL 3.0 X 10⁴ L = 273 = 2.7 X 10² g O₂ 2.73 10² g O₂ X 32.00 1 mol g O₂ O₂ 1 mol 1 mol O₂ 32.05 1 mol = 2.7 10² g 22.95 (a) + 2H₂S(aq) 3S(s) + 2H₂O(g) or, if we assume S₈ is the product, + 16H₂S(aq) 3S₈(s) + 16H₂O(g). (b) Assume that all S in the coal becomes SO₂ upon combustion, so that 1 mol S (coal) = 1 mol SO₂; 1 ton = 2000 lb; 760 torr = 1.00 atm 4000 lb coal 0.035 lb S X 453.6 g 1 mol S (coal) 1 mol 2 mol H₂S 1 lb coal 1lbS 1 mol S (coal) 1 mol SO₂ = 3960 = 4.0 10³ mol V = 3960 mol X (0.08206 - atm/mol - K) 300 K = 97,496 = 9.7 10⁴ L 1.00 atm (c) 3960 mol H₂S 2 3 mol mol H₂S X 1 mol S = 1.9 X 10⁵ gS This is about 210 lb S per ton of coal combusted. (However, two-thirds of this comes from the H₂S, which presumably at some point was also obtained from coal.) 22.96 Plan. vol air kg air g g FeS. Use the ideal-gas equation to change volume of air to mass of air, (assuming 1.00 atm, 298 K and an average molar mass (MM) for air of 29.0 g/mol. Use (20 mg H₂S/kg) air to find the mass of in the given mass of air. Solve. = 12 ft 20 ft 8 ft 12³ X 2.54³ 1³ in³ cm³ X 1000 1L 3 = 5.4368 10⁴ = 5x 10⁴ L PV MM gair = ; assume P = 1.00 atm, T = 298 K, MMair = 29.0 g/mol RT = 1.00 atm 5.4368 X 10⁴ L 29.0 g/mol mol- K = 64,476 = 6 10⁴ g air 298 K 0.08206 L - atm 6.4476 10⁴ g air 1000 1kg 20 1 mg kg air H₂S 1000 mg = 1.2895 = 1 FeS(s) + 2HCl(aq) 1.2895 g H₂ 34.08 1 mol g H₂ H₂S 1 1 mol mol H₂S FeS 87.91 1 mol FeS = 3.3263 = 3 FeS 688