Question

Suppose 316.0 g aluminium sulphide reacts with 493.0 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. 309.0 g
• C. 109.8 g
• D. 340.3 g

Answer 1

The correct option is A 265.4 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 $A{l}_2{S}_3=\frac{316.0g}{150g/mol}=2.107mol$
Number of moles of $H_{2}O=\frac{493.0g}{18g/mol}=27.388mol$

For $A{l}_2{S}_3$
$\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{2.107}{1}=2.1044$

For $H_{2}O$ $\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{27.388}{6}=4.564$

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

To find the number of moles of $H_{2}O$ required to consume 2.107 mol of $A{l}_2{S}_3$:

$\frac{(2.107molofA{l}_2{S}_3\times 6molof{H}_{2}O)}{\left(1molofA{l}_2{S}_3\right)}=12.642molof{H}_{2}O$

Weight of $H_{2}O=12.642mol\times 18g/mol=227.556g$

Number of moles of excess reagent left = 493.0 g – 227.556 = 265.444 g