JEE Main 2026 — 21 January, Morning Shift — Question 53

Reaction: $\mathrm{Ca + 2HCl \rightarrow CaCl_2 + H_2}$

Moles of calcium $=\frac{14.0}{40} = 0.35\,\mathrm{mol}$

From stoichiometry, $1\,\mathrm{mol\ Ca} \rightarrow 1\,\mathrm{mol\ H_2}$

Hence, $\mathrm{H_2}$ formed $= 0.35\,\mathrm{mol}$

Volume of $\mathrm{H_2}$ at $273\,\mathrm{K}$ and $1\,\mathrm{atm}$, $V = 0.35 \times 22.4 = 7.84\,\mathrm{L}$

Molar mass of $\mathrm{CaCl_2}$ $= 40 + 2 \times 35.5 = 111\,\mathrm{g\,mol^{-1}}$

Mass of $\mathrm{CaCl_2}$ formed $= 0.35 \times 111 = 38.85\,\mathrm{g}$

Thus statement (C) is incorrect.

Since $\mathrm{HCl}$ is in excess, calcium is the limiting reagent.