Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 113: 3.11d


$Zn(OH)_2(s)+2HNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+2H_2O(l)$

Work Step by Step

$Zn(OH)_2(s)+HNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+H_2O(l)$ $Zn$ is in balance. There are 2 $N$ atoms on the right. Add a coefficient of 2 to $HNO_3$ to balance $N$. $Zn(OH)_2(s)+2HNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+H_2O(l)$ This leaves us with $Zn$ and $N$ balanced, but there are 8 $O$ atoms on the left and 7 on the right. There are 4 $H$ atoms on the left and 2 on the right. Add a coefficient of 2 to $H_2O$ to complete the balancing of the equation. $Zn(OH)_2(s)+2HNO_3(aq)\rightarrow Zn(NO_3)_2(aq)+2H_2O(l)$
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