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: 3.39

Answer

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 $\%O=\frac{2(16.00)}{150.7}\times 100\%=21.23\%$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.