Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 5 - Gases - Questions & Problems - Page 218: 5.72


Please see the work below.

Work Step by Step

We know that $PV=nRT$ This can be rearranged as: $n=\frac{PV}{RT}$ We plug in the known values to obtain: $n=\frac{95849\times 0.0078}{8.31\times 298}$ $n=0.302 \space mol$ From the given equation, we obtain: Mole ratio ($Zn$:$H_2) =1:1$ Thus $mass=moles\times molar\space mass \space of Zn$ $mass=0.302\times 65.39=19.8g$
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