HW10 (solutions) - myers(nmm698 H10 Gases 2 mccord(50970...

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myers (nmm698) – H10: Gases 2 – mccord – (50970) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the mass of oxygen gas in a 18.0 L container at 33.0 C and 5.76 atm? Correct answer: 132 . 063 g. Explanation: T = 33 . 0 C + 273 = 306 K P = 5 . 76 atm V = 18 L m = ? n = P V R T = (5 . 76 atm)(18 L) ( 0 . 0821 L · atm mol · K ) (306 K) = 4 . 12696 mol O 2 m = (4 . 12696 mol) parenleftbigg 32 g mol parenrightbigg = 132 . 063 g O 2 002 10.0 points Toluene (C 6 H 5 CH 3 ) is a liquid compound similar to benzene (C 6 H 6 ). Calculate the mole fraction of toluene in the solution that contains 38.5 g toluene and 85.0 g benzene. Correct answer: 0 . 277. Explanation: m toluene = 38.5 g m benzene = 85.0 g n toulene = (38 . 5 g toluene) parenleftBig 1 mol 92 . 14 g parenrightBig = 0 . 418 mol n benzene = (85 . 0 g benzene) parenleftBig 1 mol 78 . 11 g parenrightBig = 1 . 09 mol The total number of moles of all species present is 0 . 418 mol + 1 . 09 mol = 1 . 51 mol The mole fraction of toluene is then X toluene = n toluene n total = 0 . 418 mol 1 . 51 mol = 0 . 277 003 (part 1 of 4) 10.0 points Iron pyrite (FeS 2 ) is the form in which much of the sulfur exists in coal. In the combustion of coal, oxygen reacts with iron pyrite to produce iron(III) oxide and sulfur dioxide, which is a major source of air pollution and a substantial contributor to acid rain. What mass of Fe 2 O 3 is produced from the reaction is 62 L of oxygen at 2 . 76 atm and 145 C with an excess of iron pyrite? Correct answer: 144 . 845 g. Explanation: P = 2 . 76 atm T = 145 C + 273 = 418 K R = 0 . 08206 L · atm K · mol V = 62 L MW Fe 2 O 3 = 2(55 . 845 g / mol) + 3(15 . 9994 g / mol) = 159 . 688 g / mol The balanced equation is 4 FeS 2 (s) + 11 O 2 (g) -→ 2 Fe 2 O 3 (s) + 8 SO 2 (g) Applying the ideal gas law to the O 2 , P V = n R T n = P V R T = (2 . 76 atm) (62 L) ( 0 . 08206 L · atm K · mol ) (418 K) = 4 . 98876 mol . From stoichiometry and the molar mass of Fe 2 O 3 , m Fe 2 O 3 = (159 . 688 g / mol Fe 2 O 3 ) × 2 mol Fe 2 O 3 11 mol O 2 (4 . 98876 mol O 2 ) = 144 . 845 g Fe 2 O 3 . 004 (part 2 of 4) 10.0 points If the sulfur dioxide that is generated above is dissolved to form 3 . 5 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 1 . 03663 M.
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myers (nmm698) – H10: Gases 2 – mccord – (50970) 2 Explanation: V = 3 . 5 L SO 2 (g) + H 2 O( ) -→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (4 . 98876 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 3 . 62819 mol . 3 . 62819 mol of SO 2 will dissolve in 3 . 5 L of water to form a solution that is 3 . 62819 mol 3 . 5 L = 1 . 03663 M in H 2 SO 4 . 005 (part 3 of 4) 10.0 points What mass of SO 2 is produced in the burning of 1 tonne (1 t = 1000 kg) of high-sulfur coal, if the coal is 4% pyrite by mass? Correct answer: 42 . 7181 kg. Explanation: m coal = 1000 kg m FeS 2 = 4%(1000 kg) = 40 kg = 40000 g MW FeS 2 = 55 . 845 g / mol + 2(32 . 065 g / mol) = 119 . 975 g MW SO 2 = 32 . 065 g / mol + 2(15 . 9994 g / mol) = 64 . 0638 g m SO 2 = 1000 kg coal parenleftbigg 40000 g FeS 2 1000 kg coal parenrightbigg × parenleftbigg 1 mol FeS 2 119 . 975 g FeS 2 parenrightbigg × parenleftbigg 8 mol SO 2 4 mol FeS 2 parenrightbigg parenleftbigg 64 . 0638 g SO 2 1 mol SO 2 parenrightbigg = 42718 . 1 g = 42 . 7181 kg SO 2 .
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