Stoichiometry of CO-NO reaction in catalytic converter

Steps:

• Balanced equation: 2CO + 2NO → 2CO₂ + N₂ (NO in excess, so CO limiting).
• Moles of CO: 5.0 g / 28 g/mol = 0.179 mol.
• Moles CO₂ produced = 0.179 mol; mass = 0.179 × 44 g/mol = 7.9 g (≈8 g).
• Moles N₂ produced = 0.179 / 2 = 0.089 mol; mass = 0.089 × 28 g/mol = 2.5 g (≈2.4 g).

Why D is correct:

• Matches stoichiometric masses: 1:1 CO:CO₂ ratio and 2:1 CO:N₂ ratio from balanced equation.

Why the others are wrong:

• A: N₂ mass too high (ignores 2:1 CO:N₂ ratio).
• B: Lists NO as product (NO is reactant, not produced).
• C: Lists NO as product and underestimates CO₂ mass.

Final answer: D