HW 6 - ching(yyc99 – Homework 6 – Lyon –(53750 1 This...

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: ching (yyc99) – Homework 6 – Lyon – (53750) 1 This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 5.0 L flask containing O 2 at 2.00 atm is connected to a 3.0 L flask containing H 2 at 4.00 atm and the gases are allowed to mix. What is the mole fraction of H 2 ? 1. 0.67 2. 0.55 correct 3. 0.45 4. Cannot be determined 5. 0.33 6. 0.25 Explanation: Applying the ideal gas law P V = nRT = P V RT N O 2 = (2 atm) (5 L) RT N H 2 = (4 atm) (3 L) RT Assume the temperature of the two gases remains the same before and after the mixing occurs. The mole fraction of H 2 is χ H 2 = n H 2 n total = n H 2 n H 2 + n O 2 = (4 atm) (3 L) RT (4 atm) (3 L) RT + (2 atm) (5 L) RT = (4 atm) (3 L) (4 atm) (3 L) + (2 atm) (5 L) = 0 . 545455 002 10.0 points Calculate the amount of oxygen gas collected by the displacement of water at 14 ◦ C if the atmospheric pressure is 790 Torr and the vol- ume is 5 L. The vapor pressure of water at 14 ◦ C is 12 Torr. 1. 0.217 mol correct 2. 0.0184 mol 3. 0.224 mol 4. 4.46 mol 5. 0.00335 mol Explanation: T = 14 ◦ C + 273.15 = 287 . 15 K P = 790 Torr V = 5 L Using Dalton’s Law, P tot = P water + P oxygen P oxygen = P tot- P water = 790 Torr- 12 Torr = 778 Torr 1 atm 760 Torr = 1 . 02368 atm The ideal gas law is P V = nRT n = P V RT = (1 . 02368 atm)(5 L) ( . 08206 L · atm mol · K ) (287 . 15 K) = 0 . 217218 mol oxygen 003 10.0 points The main composition of dry air at sea level is about 75.5% nitrogen, 23.1% oxygen, and 1.3% argon. In a 1.00 g sample of dry air at 1.00 atm, calculate the partial pressure of argon gas. 1. 9 . 43 × 10 − 3 atm correct 2. 10 . 2 × 10 − 3 atm 3. 8 . 33 × 10 − 3 atm ching (yyc99) – Homework 6 – Lyon – (53750) 2 4. 5 . 45 × 10 − 3 atm 5. 4 . 17 × 10 − 3 atm Explanation: Assume you have 100 g of this mixture; calculate the number of moles: n N 2 = (75 . 5 g N 2 ) 1 mol N 2 31 . 9988 g N 2 = 2 . 69514 mol N 2 . n O 2 = (23 . 1 g O 2 ) 1 mol O 2 31 . 9988 g O 2 = 0 . 721902 mol O 2 . n Ar = (1 . 3 g Ar) 1 mol Ar 4 . 0026 g Ar = 0 . 0325424 mol Ar . n tot = n N 2 + n O 2 + n Ar = 2 . 69514 mol N 2 + 0 . 721902 mol O 2 + 0 . 0325424 mol Ar = 3 . 44958 mol gas Dalton’s Law: P O 2 = P tot × X O 2 = P tot × n Ar 2 n tot = (1 atm) . 0325424 mol Ar 3 . 44958 mol gas = 0 . 00943372 atm . 004 10.0 points Find the number of moles of ammonia pro- duced from a reaction of 4 moles of N 2 gas at STP with hydrogen gas. 1. 2 mol 2. 4 mol 3. 8 mol correct 4. 1 mol Explanation: n = 4 mol The balanced chemical equation is N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) ? mol NH 3 = 4 mol N 2 × 2 mol NH 3 1 mol N 2 = 8 mol NH 3 005 10.0 points If sufficient acid is used to react completely with 21.0 grams of Mg Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) what volume of hydrogen at STP would be produced?...
View Full Document

This note was uploaded on 04/01/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

Page1 / 14

HW 6 - ching(yyc99 – Homework 6 – Lyon –(53750 1 This...

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