For the given reaction; $\mathrm{CaCO}_{3}+2\mathrm{HCl}\rightarrow\mathrm{CaCl}_{2}+\mathrm{H}_{2}\mathrm{O}+\mathrm{CO}_{2}$ If $90\,\mathrm{g}\,\mathrm{CaCO}_{3}$ is added to 300 mL of HCl which contains $38.55\%$ HCl by mass and has density $1.13\,\mathrm{g}\,\mathrm{mL}^{-1}$, then which of the following option is correct? Given molar mass of $\mathrm{H}$, $\mathrm{Cl}$, $\mathrm{Ca}$ and O are 1, 35.5, 40 and $16\,\mathrm{g}\,\mathrm{mol}^{-1}$ respectively.

(A) 64.97 g of HCl remains unreacted

(B) $32.85\,\mathrm{g}$ of $\mathrm{CaCO}_{3}$ remains unreacted

(D) 60.32 g of HCl remains unreacted

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Some Basic Concepts of Chemistry

For the given reaction;

CaCO_{3} + 2 HCl \to CaCl_{2} + H_{2} O + CO_{2}

90 g CaCO_{3}

is added to 300 mL of HCl which contains

38.55%

HCl by mass and has density

1.13 g mL^{- 1}

, then which of the following option is correct?

Given molar mass of

H

Cl

Ca

and O are 1, 35.5, 40 and

16 g mol^{- 1}

respectively.

Answer: (A)

The HCl solution weighs

300 \times 1.13 = 339 g

, of which

339 \times 0.3855 = 130.68 g

is HCl, i.e.

\frac{130.68}{36.5} = 3.58 mol

. The

\frac{90}{100} = 0.90 mol

CaC O_{3}

needs

2 \times 0.90 = 1.80 mol

of HCl, so

CaC O_{3}

is the limiting reactant and HCl is in excess. The HCl left over is

3.58 - 1.80 = 1.78 mol

, equal to

1.78 \times 36.5 = 64.97 g

of HCl remaining unreacted.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option