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: 3.48

Answer

Please see the work below.

Work Step by Step

We know that $\%F=\frac{mass\space of F\space in \space 1mol\space SnF_2}{molar\space mass\space of SnF_2}\times 100\%=\frac{2(19.00g)}{156.7g}\times \%=24.25\%$ Now: $(0.2425)(24.6)=5.97g\space F$
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