Iron is made by reduction of iron oxide using carbon monoxide.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
50 g of iron is made in a reaction from 80 g of iron oxide.
What is the percentage yield of iron in this reaction? (Mr Fe₂O₃ = 160, Ar Fe = 56)
A) [(50/56)/(80/160)] × 100
B) (50/80) × 100
C) (25/80) × 100
D) [(25/112)/(80/160)] × 100
E) (50/56) × 100
IMAT 2023 · Question 46
Q46 - Stoichiometric calculations
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The correct answer is E) (50/56) × 100.
80 g of Fe₂O₃ = 80/160 = 0.5 mol
Each Fe₂O₃ gives 2 Fe, so 1 mol of Fe = 56 g in theory
You got 50 g, so the yield is (50/56) × 100, about 89%.
A is the one to watch: it compares moles of Fe with moles of Fe₂O₃ but forgets the 1:2 ratio, which gives a yield above 100%.
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