H10- Gases 2-solutions

# H10- Gases 2-solutions - goldstein(ddg625 – H10 Gases 2...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: goldstein (ddg625) – 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 11.0 L container at 23.0 ◦ C and 3.88 atm? Correct answer: 56 . 2004 g. Explanation: T = 23 . ◦ C + 273 = 296 K P = 3 . 88 atm V = 11 L m = ? n = P V RT = (3 . 88 atm)(11 L) ( . 0821 L · atm mol · K ) (296 K) = 1 . 75626 mol O 2 m = (1 . 75626 mol) parenleftbigg 32 g mol parenrightbigg = 56 . 2004 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 126 g toluene and 64.0 g benzene. Correct answer: 0 . 626. Explanation: m toluene = 126 g m benzene = 64.0 g n toulene = (126 g toluene) parenleftBig 1 mol 92 . 14 g parenrightBig = 1 . 37 mol n benzene = (64 . 0 g benzene) parenleftBig 1 mol 78 . 11 g parenrightBig = 0 . 819 mol The total number of moles of all species present is 1 . 37 mol + 0 . 819 mol = 2 . 19 mol The mole fraction of toluene is then X toluene = n toluene n total = 1 . 37 mol 2 . 19 mol = 0 . 626 003 (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 65 L of oxygen at 4 . 28 atm and 164 ◦ C with an excess of iron pyrite? Correct answer: 225 . 245 g. Explanation: P = 4 . 28 atm T = 164 ◦ C + 273 = 437 K R = 0 . 08206 L · atm K · mol V = 65 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 = (4 . 28 atm) (65 L) ( . 08206 L · atm K · mol ) (437 K) = 7 . 7579 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 (7 . 7579 mol O 2 ) = 225 . 245 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 7 . 4 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 762447 M. goldstein (ddg625) – H10: Gases 2 – mccord – (50970) 2 Explanation: V = 7 . 4 L SO 2 (g) + H 2 O( ℓ )-→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (7 . 7579 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 5 . 64211 mol ....
View Full Document

## This note was uploaded on 11/11/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

### Page1 / 7

H10- Gases 2-solutions - goldstein(ddg625 – H10 Gases 2...

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

View Full Document
Ask a homework question - tutors are online