If 20.0 g of sulfur dioxide (SO₂) reacts with 10.0 g of oxygen (O₂) to produce sulfur trioxide (SO₃) according to the balanced equation 2SO₂(g) + O₂(g) → 2SO₃(g), what is the mass of the excess reactant remaining unreacted? (Molar masses: SO₂ = 64 g/mol, O₂ = 32 g/mol)

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• A 0.00 g
• B 2.50 g
• C 5.00 g
• D 7.50 g

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Answer: C. 5.00 g

Moles: SO2 = 20/64 = 0.3125, O2 = 10/32 = 0.3125. SO2 needs half its moles of O2, i.e. 0.156 mol = 0.156 x 32 = 5 g. SO2 is used up, and 10 - 5 = 5.00 g of O2 is left over.

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