H09 Gas Laws 2

# H09 Gas Laws 2 - saldana (avs387) – H09: Gas Laws 2 –...

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Unformatted text preview: saldana (avs387) – H09: Gas Laws 2 – Mccord – (52385) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. I’ve included some more gas stoichiome- try problems on this assignment. Practicing all aspects of stoichiometry is always a good thing in a chemistry class. Please get all sto- ichiometry related topics down pat. Chem- istry has stoichiometry at the heart of it. Learn it - live it. 001 10.0 points What is the mass of oxygen gas in a 10.3 L container at 13.0 ◦ C and 7.84 atm? Correct answer: 110 . 051 g. Explanation: T = 13 . ◦ C + 273 = 286 K P = 7 . 84 atm V = 10 . 3 L m = ? n = P V RT = (7 . 84 atm)(10 . 3 L) ( . 0821 L · atm mol · K ) (286 K) = 3 . 43909 mol O 2 m = (3 . 43909 mol) parenleftbigg 32 g mol parenrightbigg = 110 . 051 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 75.9 g toluene and 62.0 g benzene. Correct answer: 0 . 509. Explanation: m toluene = 75.9 g m benzene = 62.0 g n toulene = (75 . 9 g toluene) parenleftBig 1 mol 92 . 14 g parenrightBig = 0 . 824 mol n benzene = (62 . 0 g benzene) parenleftBig 1 mol 78 . 11 g parenrightBig = 0 . 794 mol The total number of moles of all species present is . 824 mol + 0 . 794 mol = 1 . 62 mol The mole fraction of toluene is then X toluene = n toluene n total = . 824 mol 1 . 62 mol = 0 . 509 003 (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 84 L of oxygen at 3 . 99 atm and 167 ◦ C with an excess of iron pyrite? Correct answer: 269 . 512 g. Explanation: P = 3 . 99 atm T = 167 ◦ C + 273 = 440 K R = 0 . 08206 L · atm K · mol V = 84 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 . 99 atm) (84 L) ( . 08206 L · atm K · mol ) (440 K) = 9 . 28256 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 (9 . 28256 mol O 2 ) = 269 . 512 g Fe 2 O 3 . saldana (avs387) – H09: Gas Laws 2 – Mccord – (52385) 2 004 (part 2 of 4) 10.0 points If the sulfur dioxide that is generated above is dissolved to form 5 . 1 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution?...
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## This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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H09 Gas Laws 2 - saldana (avs387) – H09: Gas Laws 2 –...

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