Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises: 68

Answer

$23.0\text{ L}$

Work Step by Step

$120*0.5=62.5\text{ grams $H_2O_2$}$ $62.5\text{ grams $H_2O_2$}*\frac{1\text{ mol $H_2O_2$}}{34.0147\text{ grams $H_2O_2$}}*\frac{1 \text{ mol $O_2$}}{2\text{ mol $H_2O_2$}}=0.919\text{ mol $O_2$}$ $746\text{ torr}*\frac{1\text{ atm}}{760\text{ torr}}=0.982\text{ atm}$ $PV=nRT$ $0.982*V=0.919*0.08206*300$ $V=23.0\text{ L}$
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