341 15 atm 5118 51 atm The partial pressure for each gas is C H 4 55 atmC 2 H 6

# 341 15 atm 5118 51 atm the partial pressure for each

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= .341 1.5 atm = .5118 .51 atm The partial pressure for each gas is: C H 4 = .55 atmC 2 H 6 = .44 atmC 3 H 8 = .51 atm (Reference: Chang 5.67) 11. Propane (C 3 H 8 ) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.45 g of propane. (10 points) A) C 3 H 8 ( g ) + 5 O 2 ( g )− 3 CO 2 ( g ) + 4 H 2 O ( g ) B) molC 3 H 8 = 7.45 44.09 = .169 molC 3 H 8 molC O 2 = .169 molC 3 H 8 3 molCO 2 1 molC 3 H 8 = .507 molC O 2 STP=1mol CO2 * 22.4L .507 molC O 2 22.4 = 11.35 L→ 11.4 L 11.4L of CO2 produced at STP from 7.45g of propane (Reference: Chang 5.109) 12. A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric acid at STP condition. They react as follows: 2Al + 6HCl 3H 2 ( g ) + 2AlCl 3 Calculate the following: Mol HCl= .54 V HCl= 75mL mol Al = 10 26.982 g / mo l 1 = 0.371 mol mol HCl = .54 .075 L = .0405 mol 2mol Al=3mol H2 .371mol Al= .556mol H2 6mol HCl=3mol H2 .0405mol HCl= .0203mol H2 (HCl limiting Reactant) a. Volume, in liters, of hydrogen gas. (5 points) 4 Copyright © 2017 by Thomas Edison State University . All rights reserved. V = nRT P .0203 ( .0821 ) ( 273 ) 1 = .454 L b. Molarity of Al +3 . (Assume 75.0 mL solution.) (5 points) 6 molHCl = 2 molAlC l 3 .0405 mol = .0135 mol 3 + ¿ = 0.0135 .075 L = .180 mol A l ¿ c. Molarity of Cl . (Assume 75.0 mL solution.) (5 points) ¿ .0405 .075 L = .540 mol 3 + ¿ = 3 Cl ¿ 1 mol Al ¿ 5 Copyright © 2017 by Thomas Edison State University . All rights reserved. #### You've reached the end of your free preview.

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