Stoichiometry of citric acid neutralization by sodium carbonate Steps:

• Convert volume to dm³: 20 cm³ = 0.020 dm³.
• Calculate moles of Na₂CO₃: 0.020 dm³ × 0.50 mol dm⁻³ = 0.010 mol.
• Na₂CO₃ neutralizes 2 mol H⁺ per mol (Na₂CO₃ + 2H⁺ → 2Na⁺ + H₂O + CO₂), so total H⁺ = 2 × 0.010 = 0.020 mol.
• Citric acid formula shows 3 COOH groups, providing 3 H⁺ per molecule, so moles citric acid = 0.020 / 3 ≈ 0.0067 mol. Not enough information: Exact number of H⁺ reacting (all 3 or fewer?) is ambiguous without specified titration conditions. Why B is correct: N/A—calculation yields ~0.007 mol, not matching any option precisely; B closest if assuming diprotic (0.010 mol, but still not 0.04). Why the others are wrong:
• A: Assumes monoprotic citric acid (0.020 mol H⁺ = 0.020 mol acid).
• C: Assumes ~0.10 mol Na₂CO₃ (e.g., 200 cm³ volume).
• D: Assumes ~0.18 mol Na₂CO₃ (e.g., 360 cm³ volume).

Final answer: Not enough information.