Question

225 g of aluminium sulphide reacts with 495 g of water. How much of the excess reactant is left over at the end of the reaction?
$A{l}_2{S}_3+H_{2}O\to Al(OH{)}_3+H_{2}S$

• A. 265.4 g
• B. 387 g
• C. 226 g
• D. 333 g

Answer 1

The correct option is D 333 g

The unbalanced equation is:
$A{l}_2{S}_3+H_{2}O\to Al(OH{)}_3+H_{2}S$
The balanced equation is:
$A{l}_2{S}_3+6H_{2}O\to 2Al(OH{)}_3+3H_{2}S$
Number of moles of $A{l}_2{S}_3=\frac{225g}{150g/mol}=1.5mol$
Number of moles of $H_{2}O=\frac{495g}{18g/mol}=27.5mol$

For $A{l}_2{S}_3$
$\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{1.5}{1}=1.5$

For $H_{2}O$ $\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{27.5}{6}=4.58$

$A{l}_2{S}_3$ is the limiting reagent.

To find the number of moles of $H_{2}O$ required to consume 1.5 mol of $A{l}_2{S}_3$:

$\frac{(1.5molofA{l}_2{S}_3\times 6molof{H}_{2}O)}{\left(1molofA{l}_2{S}_3\right)}=9molof{H}_{2}O$

Weight of $H_{2}O=9mol\times 18g/mol=162g$

Weight of excess reagent left = 495 g – 162 = 333 g