Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2026 — 22 January, Morning Shift — Question 54

In the reaction, $\mathrm{2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^{-}(aq) + 3H_2(g)}$ . Which of the following statement is correct?

Option A:
$11.2\,\mathrm{L}$ $\mathrm{H_2(g)}$ at STP is produced for every mole of $\mathrm{HCl}$ consumed.
Correct
Option B:
$67.2\,\mathrm{L}$ $\mathrm{H_2(g)}$ at STP is produced for every mole of $\mathrm{Al}$ that reacts.
Option C:
$12\,\mathrm{L}$ $\mathrm{HCl(aq)}$ is consumed for every $6\,\mathrm{L}$ $\mathrm{H_2(g)}$ produced.
Option D:
$33.6\,\mathrm{L}$ $\mathrm{H_2(g)}$ is produced regardless of temperature and pressure for every mole of $\mathrm{Al}$ that reacts.

Answer: A

From the balanced equation: $\mathrm{2Al + 6HCl \rightarrow 2Al^{3+} + 6Cl^- + 3H_2}$

Mole ratios, we have

$1\,\mathrm{Al} \rightarrow \frac{3}{2}\,\mathrm{H_2}$

$1\,\mathrm{HCl} \rightarrow \frac{1}{2}\,\mathrm{H_2}$

At STP, $1\,\mathrm{mol\ gas} = 22.4\,\mathrm{L}$

Volume of $\mathrm{H_2}$ per mole $\mathrm{HCl}$ $= \frac{1}{2} \times 22.4 = 11.2\,\mathrm{L}$

Thus option (A) is correct.

Volume of $\mathrm{H_2}$ per mole $\mathrm{Al}$ $= \frac{3}{2} \times 22.4 = 33.6\,\mathrm{L}$

Hence option (B) is incorrect.

Option (C) is invalid because liquid volume of $\mathrm{HCl}$ cannot be directly compared with gas volume.

Option (D) is incorrect since gas volume depends on temperature and pressure.

Answer key and solution verified before publishing.

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