In the reaction, $2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})+6 \mathrm{Cl}^{-}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g})$

(A) $11.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})$ at STP is produced for every mole of HCl consumed.

(B) $67.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})$ at STP is produced for every mole of Al that reacts.

(C) $12 \mathrm{~L} \mathrm{HCl}_{(\mathrm{aq})}$ is consumed for every $6 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})$ produced.

(D) $33.6 \quad \mathrm{~L}_{2} \quad \mathrm{H}_{2}(\mathrm{~g})$ is produced regardless of temperature and pressure for every mole of Al that reacts.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

WhatsApp Email

In the reaction,

2 Al (s) + 6 HCl (aq) \to 2 Al^{3 +} (aq) + 6 Cl^{-} (aq) + 3 H_{2} (g)

(A) $11.2 L H_{2} (g)$ at STP is produced for every mole of HCl consumed.✓ Correct
(B) $67.2 L H_{2} (g)$ at STP is produced for every mole of Al that reacts.
(C) $12 L HCl_{(aq)}$ is consumed for every $6 L H_{2} (g)$ produced.
(D) $33.6 L_{2} H_{2} (g)$ is produced regardless of temperature and pressure for every mole of Al that reacts.

Answer: (A)

Solution

The equation shows that 6 mol of HCl give 3 mol of hydrogen and that 2 mol of aluminium give 3 mol of hydrogen. For every mole of HCl consumed the hydrogen formed is

\frac{3}{6} = 0.5 mol

, which at STP occupies

0.5 \times 22.4 = 11.2 L

. Each mole of aluminium yields

1.5 mol = 33.6 L

, not

67.2

L; the volume of an aqueous reagent cannot be equated to a gas volume; and any gas volume depends on temperature and pressure. The correct statement is that 11.2 L of hydrogen at STP is produced for every mole of HCl consumed.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Some Basic Concepts of Chemistry →

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

Solution

More JEE Main questions on Some Basic Concepts of Chemistry