Stoichiometry of CO-NO reaction in catalytic converter

Steps:

• Balanced equation: 2CO + 2NO → 2CO₂ + N₂ (excess NO ensures complete CO reaction).
• Moles of CO = 5 g / 28 g/mol = 0.1786 mol.
• Moles CO₂ = 0.1786 mol; mass CO₂ = 0.1786 × 44 g/mol ≈ 7.86 g ≈ 8 g.
• Moles N₂ = 0.1786 / 2 = 0.0893 mol; mass N₂ = 0.0893 × 28 g/mol ≈ 2.5 g ≈ 2 g.

Why D is correct:

• Matches stoichiometric masses from balanced equation, using standard molar masses (CO=28, CO₂=44, N₂=28) and approximation for given option.

Why the others are wrong:

• A: Produces NO (reactant, not product) with masses from incorrect equation.
• B: CO₂ mass correct, but 9.2 g N₂ implies ~10.4 g CO (violates given 5 g).
• C: 2.4 g CO₂ implies ~1.5 g CO (wrong stoichiometry); 9.2 g N₂ too high.

Final answer: D