# 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

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.