Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 5 - Problems: 5.105

Answer

983 L

Work Step by Step

2C8H18 + 25O2 -----------> 16CO2(g) + 18H2O(g) balanced equation shows that 2 moles C8H18 produces 16 mole CO2 and 18 moles H2O (a total of 34 moles gas) So 1 mole C8H18 produces a total of 17 moles of gas moles C8H18 = mass / molar mass = 125 g / 114.224 g/mol = 1.094 moles Therefore moles gas products = 17 x 1.094 mol = 18.6 moles of gaseous products. V = 18.6 mol x 62.36367 torrLmol^-1K^-1 x 623.15 K / 735 torr = 983 L b) 2 moles C8H18 requires 25 moles of O2 to fully react So 1.094 moles C8H18 needs (25/2 x 1.094) moles O2 = 13.675 moles O2 needed Volume of this many moles of O2 under the given conditions is V = 13.675 mol x 62.36367 x 623.15 K / 735 torr = 723 L O2 is 21.0 % of air, thus 723 L is 21.0 % of the total volume of air needed 21.0 / 100 x volume air = 723 L Therefore volume air needed = 723 L / 0.21 = 3443 L Air in exhaust = total air - O2 consumed = 3443 L - 723 L = 2720 L Total gases in exhaust = unused air + gaseous products = 2720 L + 983 L = 3703 L
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