Question

Suppose 316.0 g of aluminium sulphide reacts with 493.0 g of water, what mass of the excess reactant will remain?
Given reaction: $A{l}_2{S}_3+6H_{2}O\to 2Al(OH{)}_3+3H_{2}S$

• A. 265.14 g
• B. 108.52 g
• C. 400 g
• D. 66.25 g

Answer 1

The correct option is A 265.14 g

The unbalanced equation is:
$A{l}_2{S}_3+H_{2}O\to Al(OH{)}_3+H_{2}S$
On balancing the equation,
$A{l}_2{S}_3+6H_{2}O\to 2Al(OH{)}_3+3H_{2}S$
Moles of $A{l}_2{S}_3=\frac{316}{150}$ =2.11 moles
Moles of water $=\frac{493}{18}=27.39$ moles
1 mole of $A{l}_2{S}_3$ reacts with 6 moles of water.
2.11 moles of $A{l}_2{S}_3$ reacts with $\frac{6}{1}\times 2.11=12.66$ moles of water
So, $A{l}_2{S}_3$ is the limiting reagent.
Mass of excess reagent left = $(27.39-12.66)$ mole $\times \frac{18g}{mole}$
Mass of excess reagent left = 265.14 g