The atomic weights of the constituent elements are Al=26.98154, Fe=55.845, O=15.9994 amu.
We can work out the molecular weight of Fe₂O₃=2×55.845+3×15.9994=159.6882.
Then we find the number N=weight in grams/molecular or atomic weight:
For Al we get N=124/26.98154≃4.596 and for Fe₂O₃, N=601/159.6882≃3.764.
N for Al is bigger than N for Fe₂O₃, so we appear to have an excess of Al. If we multiply the atomic weight of Al by 3.764 we get 101.55 approx. So it would appear that the excess Al weighs 124-101.55=22.45g.