One of the reactions for rusting iron is as follows: 4fe + 3o2 → 2fe2o3 (mm fe: 55.85 g/mol; mm o2=32 g/mol; mm fe2o3=159.70 g/mol) if 2.80 moles of iron, fe, rust completely, how many grams of iron (iii) oxide are formed? a. 0.0351 g b. 313 g c. 894 g d. 224 g

D. 224 g

Explanation:

Let's consider the rusting of iron reaction.

4 Fe + 3 O₂ → 2 Fe₂O₃

The molar ratio of Fe to Fe₂O₃ is 4:2. The moles of Fe₂O₃ produced from 2.80 moles of Fe are:

2.80 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) = 1.40 mol Fe₂O₃.

The molar mass of Fe₂O₃ is 159.70 g/mol. The mass of Fe₂O₃ is:

1.40 mol × (159.70 g / 1 mol) = 224 g

b is the correct answer

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