Answer: 78.69 g of Fe
Explanation: Iron(III) oxide is Fe2O3. It has a molar mass of 159.7 g/mole. That consists of:
2 x Fe at 55.85 = 11.69 or 69.94%
3 x O at 16.0 = 48.0 or 30.06%
Therefore 112.5 grams will contain (112.5 g)*(69.94%) = 78.69 g of Fe
