Question

267 g of $AlC{l}_3$ reacts with 120 g of NaOH to give aluminium hydroxide. The produced aluminium hydroxide reacts with 3 mol of $H_{2}S{O}_4$ to give aluminium sulphate. Calculate the moles of aluminium sulphate produced in the reaction.
$AlC{l}_3+3NaOH\to Al(OH{)}_3+3NaCl$
$2Al(OH{)}_3+3H_{2}S{O}_4\to A{l}_2(S{O}_4{)}_3+6H_{2}O$

• A. 0.5 mol
• B. 0.7 mol
• C. 0.4 mol
• D. 1.2 mol

Answer 1

The correct option is A 0.5 mol

$AlC{l}_3+3NaOH\to Al(OH{)}_3+3NaCl$
$2Al(OH{)}_3+3H_{2}S{O}_4\to A{l}_2(S{O}_4{)}_3+6H_{2}O$

$\text{Moles of}AlC{l}_3=\frac{\text{given mass}}{\text{molar mass}}=\frac{267}{133.5}=2mol\text{Moles of NaOH}=\frac{\text{given mass}}{\text{molar mass}}=\frac{120}{40}=3mol\text{For}AlC{l}_3=\frac{\text{given moles}}{\text{Stoichiometric coefficient}}=\frac{2}{1}=2\text{For NaOH}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{3}{3}=1$

Since the ratio is lower for NaOH, it will be the limiting reagent.
$\text{3 mol of NaOH produces 1 mol of}Al(OH{)}_3$

In reaction (ii),
$\text{For}Al(OH{)}_3=\frac{\text{given moles}}{\text{Stoichiometric coefficient}}=\frac{1}{2}=0.5\text{For}{H}_{2}S{O}_4=\frac{\text{given moles}}{stoichiometriccoefficient}=\frac{3}{3}=1$
So, $Al(OH{)}_3$ will be the limiting reagent.
2 mol of $Al(OH{)}_3$ produce 1 mol of $A{l}_2(S{O}_4{)}_3$
1 mol will produce 0.5 mol of $A{l}_2(S{O}_4{)}_3$