Q14 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry
For the given reaction:
$\mathrm{CaCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
If $90 \mathrm{~g} \mathrm{CaCO}_{3}$ is added to 300 mL of HCl which contains $38.55 \% \mathrm{HCl}$ 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.
$\mathrm{CaCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
If $90 \mathrm{~g} \mathrm{CaCO}_{3}$ is added to 300 mL of HCl which contains $38.55 \% \mathrm{HCl}$ 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.
- Correct answerA64.97 g of HCl remains unreacted
- B60.32 g of HCl remains unreacted
- C$97.30 \mathrm{~g} \text { of } \mathrm{HCl} \text { reacted }$
- D32.85 g of $\mathrm{CaCO}_{3}$ remains unreacted
Solution
Molar masses: HCl = 36.5, CaCO₃ = 100 g/mol
HCl solution: mass = 300 × 1.13 = 339 g
HCl present: 339 × 0.3855 = 130.74 g = 130.74/36.5 = 3.58 mol
CaCO₃: 90/100 = 0.9 mol
Reaction: $CaCO_3 + 2HCl → CaCl_2 + H_2O + CO_2$
HCl required: 0.9 × 2 = 1.8 mol
HCl available: 3.58 mol > 1.8 mol
CaCO₃ is limiting. HCl remaining: (3.58 - 1.8) × 36.5 = 1.78 × 36.5 = 64.97 g
HCl solution: mass = 300 × 1.13 = 339 g
HCl present: 339 × 0.3855 = 130.74 g = 130.74/36.5 = 3.58 mol
CaCO₃: 90/100 = 0.9 mol
Reaction: $CaCO_3 + 2HCl → CaCl_2 + H_2O + CO_2$
HCl required: 0.9 × 2 = 1.8 mol
HCl available: 3.58 mol > 1.8 mol
CaCO₃ is limiting. HCl remaining: (3.58 - 1.8) × 36.5 = 1.78 × 36.5 = 64.97 g
Hence, the correct answer is option: 'A'.