Answer
$3H_2S(g)+2Fe(OH)_3(s)\rightarrow Fe_2S_3(s)+6H_2O(g)$
Work Step by Step
Hydrogen sulfide is $H_2S$. Iron (III) hydroxide is $Fe(OH)_3$. Iron (III) sulfide is $Fe_2S_3$ and water is $H_2O$.
$H_2S(g)+Fe(OH)_3(s)\rightarrow Fe_2S_3(s)+H_2O(g)$
Add a coefficient of 3 to $H_2S$ to balance the $S$.
Add a coefficient of 2 to $Fe(OH)_3$ to balance $Fe$.
$3H_2S(g)+2Fe(OH)_3(s)\rightarrow Fe_2S_3(s)+H_2O(g)$
Add a coefficient of 6 to $H_2O$ to balance $H$ and $O$.
$3H_2S(g)+2Fe(OH)_3(s)\rightarrow Fe_2S_3(s)+6H_2O(g)$
