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Answer: 3. 1.32 g

Key idea: find moles of HCl, use the 1 : 2 stoichiometry, then correct for purity.

  1. Moles of HCl \(=0.5\times0.050=0.025\) mol.
  2. \(\mathrm{CaCO_3}\) needed \(=0.025/2=0.0125\) mol.
  3. Mass \(=0.0125\times100=1.25\) g of pure \(\mathrm{CaCO_3}\).
  4. Impure sample \(=1.25\times100/95=1.3158\approx1.32\) g.

Option 2 (1.25 g) forgets the purity correction.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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