Limiting reactant controls CO₂ yield Steps:

• Na₂CO₃ (0.080 mol) requires 0.160 mol HCl (2:1 ratio), but only 0.100 mol HCl available, so HCl limits.
• 2 mol HCl produce 1 mol CO₂, so 0.100 mol HCl yields 0.050 mol CO₂.
• At RTP, 1 mol gas = 24 dm³, so volume = 0.050 × 24 = 1.20 dm³.

Why B is correct:

• Matches CO₂ from limiting HCl using 24 dm³/mol molar volume at RTP.

Why the others are wrong:

• A: Assumes 1:1 HCl:CO₂ ratio, ignoring stoichiometry.
• C: Uses full Na₂CO₃ (0.080 mol × 24 = 1.92 dm³), ignoring HCl limit.
• D: Doubles Na₂CO₃ volume, perhaps assuming 48 dm³/mol.

Final answer: B