NEET 2022 — Chemistry solution

NEET2022Chemistry

What mass of 95% pure CaCO $_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction ?

CaCO $_{3}$ (s) + 2HCl(aq) $\to$ CaCl $_{2}$ (aq) + CO $_{2}$ (g) + H $_{2}$ O(l)

[Calculate upto second place of decimal point]

Solution

Moles of HCl used: $n_{H C l} = 0.5 \times \frac{50}{1000} = 0.025$ mol.

Stoichiometry: 1 mole of CaCO $_{3}$ neutralises 2 moles of HCl, so moles of pure CaCO $_{3}$ needed $= \frac{0.025}{2} = 0.0125$ mol.

Mass of pure CaCO $_{3} = 0.0125 \times 100 = 1.25$ g.

The sample is only 95% pure, so mass of impure sample $= \frac{1.25}{0.95} \times 100% = \frac{1.25 \times 100}{95} \approx 1.32$ g.

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