Chemistry 12th Edition

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

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 108: 3.39


Please see the work below.

Work Step by Step

We know that: $Molar\space mass\space of \space SnO_2=118.7+2(16.00)=150.7g$ Now: $\%Sn=\frac{118.7}{150.7}\times 100\%=78.77\%$ and $\%$ composition of oxygen is calculated as follows $\%O=\frac{2(16.00)}{150.7}\times 100\%=21.23\%$
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