What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to

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Question

What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]

Accepted Answer

Correct Answer: C - 1.32 gStep-by-step Solution:Step 1 - Moles of HCl: n(HCl) = Molarity × Volume (L) = 0.5 × 0.050 = 0.025 mol.Step 2 - Moles of CaCO3 required: From stoichiometry, 1 mol CaCO3 neutralises 2 mol HCl. So moles of pure CaCO3 needed = 0.025/2 = 0.0125 mol.Step 3 - Mass of pure CaCO3: Molar mass of CaCO3 = 40 + 12 + 48 = 100 g/mol. Mass of pure CaCO3 = 0.0125 × 100 = 1.25 g.Step 4 - Mass of 95% pure sample: Mass of impure sample = (1.25 × 100)/95 = 1.3157... ≈ 1.32 g.Why other options are incorrect: Option B (1.25 g) is the mass of pure CaCO3, ignoring the 95% purity. Option A (9.50) is far too high. Option D (3.65) does not correspond to any correct calculation path.Key Concepts:Moles = Molarity × Volume (in litres). Stoichiometric ratio from balanced equation determines mole relationship.Mass of impure sample = (mass of pure substance × 100) / % purity.Molar mass of CaCO3 = 100 g/mol (Ca=40, C=12, O3=48).

What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]

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What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]

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