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: 3.13c

Answer

$2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+6H_2O(l)$

Work Step by Step

$Fe(OH)_3(s)+H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+H_2O(l)$ Add a coefficient of 2 to $Fe(OH)_3(s)$ to balance $Fe$. Add a coefficient of 3 to $H_2SO_4$ to balance S. $2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+H_2O(l)$ There are now 18 $O$ atoms on the left and 13 on the right so an additional 5 are needed. There are now 12 $H$ atoms on the left and 2 on the right, so an additional 10 are needed. Increase the coefficient on $H_2O$ from 1 to 6 to add the necessary atoms. $2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+6H_2O(l)$
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