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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2026 · 28 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

For the given reaction: CaCO_ 3 +2 HCl CaCl_ 2 +H_ 2 O+CO_ 2 If 90...

For the given reaction:
CaCO3+2HCl⟶CaCl2+H2O+CO2\mathrm{CaCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}
If 90 gCaCO390 \mathrm{~g} \mathrm{CaCO}_{3} is added to 300 mL of HCl which contains 38.55%HCl38.55 \% \mathrm{HCl} by mass and has density 1.13 g mL−11.13 \mathrm{~g} \mathrm{~mL}^{-1}, then which of the following option is correct ?
Given molar mass of H,Cl,Ca\mathrm{H}, \mathrm{Cl}, \mathrm{Ca} and O are 1, 35.5, 40 and 16 g mol−116 \mathrm{~g} \mathrm{~mol}^{-1} respectively.

Options

  1. A64.97 g of HCl remains unreacted
  2. B60.32 g of HCl remains unreacted
  3. C97.30g of HCl97.30 \mathrm{ g} \text{ of } \mathrm{HCl} reacted \text{ reacted }
  4. D32.85 g of CaCO3\mathrm{CaCO}_{3} remains unreacted

Correct answer

A. 64.97 g of HCl remains unreacted

Step-by-step 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: CaCO3+2HCl→CaCl2+H2O+CO2CaCO_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
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