Question

$0.18\text{g}$ of $Al$ reacts with $dil.HCl$ to give $H_2$. The volume of $H_2$ evolved at STP is:
(Atomic weight of $Al=27$)

• A. 1.12 litre
• B. 0.224 litre
• C. 3.33 litre
• D. 4.44 litre

Answer 1

The correct option is B 0.224 litre

$2Al\text{(s)}+6HCl\text{(aq)}\to 2AlC{l}_3\text{(s)}+3H_2\text{(g)}$
At STP, 2 mol of aluminium i.e. $(2\times 27\text{g})$ of $Al$ will produce $(3\times 22.4)\text{L}$ of $H_2$
$0.18\text{g}$ of $Al$ will produce $(\frac{3\times 22.4}{2\times 27}\times 0.18)=0.224\text{L}$ of $H_2$