Stoichiometry of NaHCO₃-H₂SO₄ reaction for percentage calculation

Steps:

• Calculate moles of H₂SO₄: 0.100 mol dm⁻³ × (15/1000) dm³ = 0.0015 mol.
• Use 1:1 mole ratio (NaHCO₃ + ½H₂SO₄ → products), so moles of NaHCO₃ = 0.0015 mol.
• Mass of NaHCO₃ = 0.0015 mol × 84 g mol⁻¹ = 0.126 g.
• Percentage by mass = (0.126 g / 1.00 g) × 100% = 12.6% ≈ 12%.

Why C is correct:

• Matches calculation using balanced equation NaHCO₃ + ½H₂SO₄ → ½Na₂SO₄ + H₂O + CO₂, where 1 mol NaHCO₃ reacts with 1 mol H₂SO₄ equivalents.

Why the others are wrong:

• A: Underestimates by factor of 4, ignoring reaction stoichiometry.
• B: Underestimates by factor of 2, possibly confusing with half the acid moles.
• D: Assumes 2:1 NaHCO₃:H₂SO₄ ratio, double-counting acid equivalents.

Final answer: C