H10- Gases 2-solutions - phan (dtp377) H10: Gases 2 mccord...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: phan (dtp377) H10: Gases 2 mccord (51600) 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 138 g toluene and 106 g benzene. Correct answer: 0 . 524. Explanation: m toluene = 138 g m benzene = 106 g n toulene = (138 g toluene) parenleftBig 1 mol 92 . 14 g parenrightBig = 1 . 49 mol n benzene = (106 g benzene) parenleftBig 1 mol 78 . 11 g parenrightBig = 1 . 36 mol The total number of moles of all species present is 1 . 49 mol + 1 . 36 mol = 2 . 85 mol The mole fraction of toluene is then X toluene = n toluene n total = 1 . 49 mol 2 . 85 mol = 0 . 524 003 (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 74 L of oxygen at 2 . 95 atm and 148 C with an excess of iron pyrite? Correct answer: 183 . 464 g. Explanation: P = 2 . 95 atm T = 148 C + 273 = 421 K R = 0 . 08206 L atm K mol V = 74 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 = (2 . 95 atm) (74 L) ( . 08206 L atm K mol ) (421 K) = 6 . 31888 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 (6 . 31888 mol O 2 ) = 183 . 464 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 6 . 6 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 696295 M. phan (dtp377) H10: Gases 2 mccord (51600) 2 Explanation: V = 6 . 6 L SO 2 (g) + H 2 O( )- H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (6 . 31888 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 4 . 59555 mol . 4 . 59555 mol of SO 2 will dissolve in 6 . 6 L of water to form a solution that is 4 . 59555 mol 6 . 6 L = 0 . 696295 M in H 2 SO 4 ....
View Full Document

Page1 / 7

H10- Gases 2-solutions - phan (dtp377) H10: Gases 2 mccord...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online