Answer
Please see the work below.
Work Step by Step
We know that
$175\space$ tablets $\times \frac{68.4\times 10^{-3}}{1 \space tablet}\times \frac{1\space mol\space CaCO_3}{100.09g\space CaCO_3}=0.1196 mol \space CaCO_3$
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
Published by
Cengage Learning
ISBN 10:
1-28505-137-8
ISBN 13:
978-1-28505-137-6
Chapter 3 - Calculations with Chemical Formulas and Equations - Questions and Problems - Page 124: 3.125
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