Moles of CO2 determine gas volume at fixed T and P

Steps:

• Identify stoichiometry: each reaction produces 1 mol CO2 per mol of carbonate or carbon source.
• A: 1g C (12 g/mol) yields 1/12 ≈ 0.083 mol CO2.
• B: 10g CaCO3 (100 g/mol) yields 10/100 = 0.1 mol CO2.
• C: 10g Na2CO3 (106 g/mol) yields 10/106 ≈ 0.094 mol CO2.
• D: 5 cm³ of 1 mol dm⁻³ H₂SO₄ = 0.005 mol acid, yields 0.005 mol CO2 (limiting reactant).

Why B is correct:

• B generates 0.1 mol CO2, the maximum, so largest volume by ideal gas law (V = nRT/P).

Why the others are wrong:

• A: 0.083 mol CO2, fewer moles than B due to lower carbon mass equivalent.
• C: 0.094 mol CO2, slightly less than B from higher molar mass of Na₂CO₃.
• D: Only 0.005 mol CO2, limited by small acid volume despite excess CuCO₃.

Final answer: B