We are given the starting amount of Al metal to be used. This will be the starting point of the calculations. We do as follows:
74.00 g ( 1 mol / 26.98 g ) ( 1 mol Al2O3 / 2 mol Al ) ( 101.96 g / 1 mol ) = 139.83 g Al2O3 produced
Hope this answers the question. Have a nice day.
