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

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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?...
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## 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.

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

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