Stoichiometry of excess Cl₂ with NaOH forming chlorate and chloride Steps:

• Moles of NaOH = 0.6 mol dm⁻³ × 1 dm³ = 0.6 mol (volume taken as 1 dm³ for calculation).
• Balanced equation: 3Cl₂ (excess) + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.
• Stoichiometric ratio: 5 mol NaCl produced per 6 mol NaOH.
• Moles of NaCl = (5/6) × 0.6 mol = 0.5 mol. Why B is correct:
• Follows the defined stoichiometry of the disproportionation reaction in hot, concentrated NaOH, where excess Cl₂ ensures complete conversion to chlorate, yielding 5/6 mol NaCl per mol NaOH. Why the others are wrong:
• A: Assumes cold dilute conditions forming hypochlorite (Cl₂ + 2NaOH → NaCl + NaOCl + H₂O), giving (1/2) × 0.6 = 0.3 mol NaCl.
• C: Assumes 1:1 NaOH to NaCl ratio, ignoring actual reaction stoichiometry.
• D: Assumes 2:1 NaCl to NaOH ratio, which overproduces NaCl beyond the balanced equation.

Final answer: B