HW07 Quy - perlick (app582) HW07 quy (50970) This print-out...

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perlick (app582) – HW07 – quy – (50970) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points What is the mass oF oxygen gas in a 17.7 L container at 29.0 C and 6.19 atm? Correct answer: 141 . 405 g. Explanation: T = 29 . 0 C + 273 = 302 K P = 6 . 19 atm V = 17 . 7 L m = ? n = P V RT = (6 . 19 atm)(17 . 7 L) ( 0 . 0821 L · atm mol · K ) (302 K) = 4 . 4189 mol O 2 m = (4 . 4189 mol) p 32 g mol P = 141 . 405 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 109 g toluene and 104 g benzene. Correct answer: 0 . 471. Explanation: m toluene = 109 g m benzene = 104 g n toulene = (109 g toluene) ± 1 mol 92 . 14 g ² = 1 . 18 mol n benzene = (104 g benzene) ± 1 mol 78 . 11 g ² = 1 . 33 mol The total number oF moles oF all species present is 1 . 18 mol + 1 . 33 mol = 2 . 52 mol The mole Fraction oF toluene is then X toluene = n toluene n total = 1 . 18 mol 2 . 52 mol = 0 . 471 003 (part 1 of 4) 10.0 points Iron pyrite (±eS 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 ±e 2 O 3 is produced From the reaction is 64 L oF oxygen at 2 . 31 atm and 158 C with an excess oF iron pyrite? Correct answer: 121 . 365 g. Explanation: P = 2 . 31 atm T = 158 C + 273 = 431 K R = 0 . 08206 L · atm K · mol V = 64 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 ±eS 2 (s) + 11 O 2 (g) -→ 2 ±e 2 O 3 (s) + 8 SO 2 (g) Applying the ideal gas law to the O 2 , P V = nRT n = P V RT = (2 . 31 atm) (64 L) ( 0 . 08206 L · atm K · mol ) (431 K) = 4 . 18007 mol . ±rom stoichiometry and the molar mass oF ±e 2 O 3 , m Fe 2 O 3 = (159 . 688 g / mol ±e 2 O 3 ) × 2 mol ±e 2 O 3 11 mol O 2 (4 . 18007 mol O 2 ) = 121 . 365 g ±e 2 O 3 . 004 (part 2 of 4) 10.0 points IF the sulFur dioxide that is generated above is dissolved to Form 7 . 2 L oF aqueous solu- tion, what is the molar concentration oF the resulting sulFurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 422229 M.
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perlick (app582) – HW07 – quy – (50970) 2 Explanation: V = 7 . 2 L SO 2 (g) + H 2 O( ) -→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (4 . 18007 mol) p 8 n SO 2 11 n O 2 P = 3 . 04005 mol . 3 . 04005 mol of SO 2 will dissolve in 7 . 2 L of water to form a solution that is 3 . 04005 mol 7 . 2 L = 0 . 422229 M in H 2 SO 4 . 005 (part 3 of 4) 10.0 points
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This note was uploaded on 12/03/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.

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HW07 Quy - perlick (app582) HW07 quy (50970) This print-out...

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