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

JEE Main 2025 — 2 April, Morning Shift — Question 16

$\mathrm{CaCO}_{3}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})$ Consider the above reaction, what

mass of $\mathrm{CaCl}_{2}$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $\mathrm{CaCO}_{3}$ ? (Given: Molar mass of

$\mathrm{Ca}, \mathrm{C}, \mathrm{O}, \mathrm{H}$ and Cl are 40, $12,16,1$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ , respectively)

Option A:
3.908 g
Option B:
2.636 g
Option C:
10.545 g
Correct
Option D:
5.272 g

Answer: C

Step-by-step solution

$\underset{10 \text { mole }}{\mathrm{CaCO}_{3}}+\underset{\mathrm{LR}}{2 \mathrm{HCl}} \rightarrow \mathrm{CaCl}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

Moles of $\mathrm{CaCl}_{2}=\frac{0.250 \times 0.76}{2}$

Mass of $\mathrm{CaCl}_{2}=\frac{0.250 \times 0.76}{2} \times 111=10.545 \mathrm{~g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Chemical Equations, Stoichiometry and Limiting Reagent

Answer: C

Step-by-step solution

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