HW10 solution_pdf

# HW10 solution_pdf - tan(jt27678 – H10 Gases 2 – mccord...

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Unformatted text preview: tan (jt27678) – 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 12.0 L container at 7.00 ◦ C and 3.44 atm? Correct answer: 57 . 463 g. Explanation: T = 7 . 00 ◦ C + 273 = 280 K P = 3 . 44 atm V = 12 L m = ? n = P V RT = (3 . 44 atm)(12 L) ( . 0821 L · atm mol · K ) (280 K) = 1 . 79572 mol O 2 m = (1 . 79572 mol) parenleftbigg 32 g mol parenrightbigg = 57 . 463 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 50.8 g toluene and 33.0 g benzene. Correct answer: 0 . 566. Explanation: m toluene = 50.8 g m benzene = 33.0 g n toulene = (50 . 8 g toluene) parenleftBig 1 mol 92 . 14 g parenrightBig = 0 . 551 mol n benzene = (33 . 0 g benzene) parenleftBig 1 mol 78 . 11 g parenrightBig = 0 . 422 mol The total number of moles of all species present is . 551 mol + 0 . 422 mol = 0 . 974 mol The mole fraction of toluene is then X toluene = n toluene n total = . 551 mol . 974 mol = 0 . 566 003 (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 86 L of oxygen at 3 . 35 atm and 158 ◦ C with an excess of iron pyrite? Correct answer: 236 . 507 g. Explanation: P = 3 . 35 atm T = 158 ◦ C + 273 = 431 K R = 0 . 08206 L · atm K · mol V = 86 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 = nRT n = P V RT = (3 . 35 atm) (86 L) ( . 08206 L · atm K · mol ) (431 K) = 8 . 14581 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 (8 . 14581 mol O 2 ) = 236 . 507 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 4 L of aqueous solution, what is the molar concentration of the result- ing sulfurous acid (H 2 SO 3 ) solution? Correct answer: 1 . 48106 M. tan (jt27678) – H10: Gases 2 – mccord – (50970) 2 Explanation: V = 4 L SO 2 (g) + H 2 O( ℓ )-→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (8 . 14581 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 5 . 92423 mol . 5 . 92423 mol of SO 2 will dissolve in 4 L of water to form a solution that is 5 . 92423 mol 4 L = 1 . 48106 M in H 2 SO 4 ....
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## This note was uploaded on 09/01/2011 for the course PHY 303 taught by Professor Erskine/tsoi during the Spring '08 term at University of Texas.

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HW10 solution_pdf - tan(jt27678 – H10 Gases 2 – mccord...

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