Limiting reactant in chlorate formation reaction

Steps:

• Calculate initial moles of NaOH: 100 cm³ = 0.1 dm³, so 0.1 × 4.0 = 0.4 mol.
• Correct balanced equation: 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O (stoichiometric ratio NaOH:Cl₂ = 2:1).
• Moles of Cl₂ required to react all NaOH: 0.4 mol NaOH ÷ 2 = 0.2 mol Cl₂ (Cl₂ coefficient is 3 in balanced equation).
• NaOH is limiting (Cl₂ bubbled until complete), so all NaOH reacts; final [NaOH] = 0 mol dm⁻³ (less than 4.0).

Why B is correct:

• Row matches Cl₂ coefficient (3) and final [NaOH] (0, less than 4.0) per limiting reactant definition.

Why the others are wrong:

• A: Assumes [NaOH] unchanged at 4.0, ignoring reaction completion.
• C: Uses NaOH coefficient (6) incorrectly for Cl₂; [NaOH] is less than 4.0.
• D: Same error as C; duplicate or formatting issue, but incorrect coefficient.

Final answer: B