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 110: 3.73


= 0.3005 mol of $H_{2}O$

Work Step by Step

$CuSO_{4}5H_{2}O$ -> $CuSO_{4}+5H_{2}O$ mol of$H_{2}O$= $\frac{(15.01gCuSO_{4}5H_{2}O)\times(1molCuSO_{4}5H_{2}O)\times(5molH_{2}O)}{(249.72gCuSO_{4}5H_{2}O)\times(1molCuSO_{4}5H_{2}O}$ = 0.3005 mol
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