Stoichiometry of hot NaOH-Cl₂ reaction limits [NaCl] Steps:

• Initial moles NaOH = 4.0 mol dm⁻³ × 0.100 dm³ = 0.40 mol.
• Correct balanced equation: 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O (Cl₂ in excess, NaOH limiting).
• Moles NaCl formed = (5/6) × 0.40 mol = 0.333 mol.
• Final [NaCl] = 0.333 mol / 0.100 dm³ ≈ 3.3 mol dm⁻³ (volume unchanged).

Why B is correct:

• Stoichiometric ratio (5 mol NaCl per 6 mol NaOH) yields [NaCl] = (5/6) × 4.0 ≈ 3.3 mol dm⁻³, less than initial 4.0 mol dm⁻³.

Why the others are wrong:

• A: [NaCl] ≈ 3.3 mol dm⁻³, not 4.0 (incomplete conversion of OH⁻ to Cl⁻).
• C: [NaCl] ≈ 3 mol dm⁻³ equivalent, not 6 (ignores 5:6 ratio).
• D: [NaCl] ≈ 3 mol dm⁻³ equivalent, not 6 (ignores 5:6 ratio).

Final answer: B