Neutralization stoichiometry of H₂SO₄ and NaOH Steps: - Balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (2 mol H₂O per mol H₂SO₄). - Moles = concentration (mol dm⁻³) × volume (dm³); for B, H₂SO₄ = 1 × 0.5 = 0.5 mol, NaOH = 1 × 1 = 1 mol. - Limiting reactant determined by min(moles H₂SO₄, moles NaOH/2); for B, min(0.5, 1/2 = 0.5) = 0.5 mol H₂SO₄. - Moles H₂O = 2 × 0.5 = 1 mol. Why B is correct: - Provides exact 1:2 mole ratio of H₂SO₄:NaOH in the balanced equation, yielding 1 mol H₂O with complete reaction. Why the others are wrong: - A: 0.25 mol H₂SO₄ and 0.5 mol NaOH yield min(0.25, 0.25) = 0.25 mol reacting units, forming 0.5 mol H₂O. - C: 0.5 mol H₂SO₄ and 0.5 mol NaOH yield min(0.5, 0.25) = 0.25 mol reacting units, forming 0.5 mol H₂O. - D: 0.5 mol H₂SO₄ and 1 mol NaOH match B, but assuming intended 1 mol H₂SO₄ and 0.5 mol NaOH, yields min(1, 0.25) = 0.25 mol reacting units, forming 0.5 …