JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

For the given reaction;

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

If 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.

A
64.97 g of HCl remains unreacted
B
32.85 g of CaCO $_{3}$ remains unreacted
C
97.30 g of HCl reacted
D
60.32 g of HCl reamains unreacted

Solution

Mass of HCl solution $= 300 mL \times 1.13 g/mL = 339$ g.

Mass of pure HCl $= 339 \times 0.3855 = 130.68$ g. Moles HCl $= \frac{130.68}{36.5} = 3.58$ mol.

Moles CaCO $_{3} = \frac{90}{100} = 0.90$ mol.

Stoichiometry needs 2 mol HCl per mol CaCO $_{3}$ , so 0.90 mol CaCO $_{3}$ needs $1.80$ mol HCl. HCl supplied (3.58 mol) is in excess; CaCO $_{3}$ is the limiting reagent and is fully consumed.

Unreacted HCl $= 3.58 - 1.80 = 1.78$ mol $= 1.78 \times 36.5 = 64.97$ g.

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