316.0 g of aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?
The unbalanced equation is
$A{l}_2{S}_3+H_{2}O\to Al(OH{)}_3+H_{2}S$

The correct option is B 265.14 g
$A{l}_2{S}_3+6H_{2}O\to 2Al(OH{)}_3+3H_{2}S$
Determining moles of the reactant:
For $A{l}_2{S}_3\Rightarrow \frac{316}{150}=2.11$ mol
For $H_{2}O\Rightarrow \frac{493}{18}=27.39$ mol
Finding limiting reagent
For $A{l}_2{S}_3=\frac{2.11}{1}=2.11$
For $H_{2}O=\frac{27.39}{6}=4.56$
$A{l}_2{S}_3$ is the limiting reagent.
1 mol of $A{l}_2{S}_3$ requires 6 mol of water
2.11 mol of $A{l}_2{S}_3$ will require $=6\times 2.11=12.66$ mol
Mass of water consumed $=12.66\times 18=227.88$ g
Mass of excess reactant left $=493–227.88=265.12$ g