Chemistry 9th Edition

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

Chapter 5 - Gases - Exercises - Page 237: 67

Answer

$135\text{ grams $NaN_3$}$

Work Step by Step

$70.0 \text{ L $N_2$}*\frac{1\text{ mol $N_2$}}{22.42\text{ L $N_2$}}*\frac{2 \text{ mol $NaN_3$}}{3 \text{ mol $N_2$}}*\frac{65.0099 \text{ grams $NaN_3$}}{1 \text{ mol $NaN_3$}}=135\text{ grams $NaN_3$}$
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