Problem 6.7: The simplified rust reaction is 4Fe 3O 2FeO To
From Chapter 6 Problem Set. Level 2: Applied Thinking.
Question
The simplified rust reaction is: 4Fe + 3O₂ → 2Fe₂O₃. To produce 10 pounds of rust (Fe₂O₃), how many pounds of iron must be destroyed? (Fe = 55.8, O = 16, Fe₂O₃ = 159.6 g/mol)
Answer
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Mass fraction of Fe in Fe₂O₃: (2 × 55.8) / 159.6 = 111.6 / 159.6 = 0.699 (≈70%). 10 lb rust × 0.699 = 6.99 lb of iron destroyed.