How much aluminum oxide and how much carbon are needed to prepare 573 g of aluminum by the balanced chemical reaction (below) if the reaction proceeds to 79.3 % yield ?

1387.2g of Al2O3

Explanation:

The number of moles of aluminum oxide required to give the actual yield is obtained from the given mass of aluminum produced by the reaction. The percentage yield is used to obtain the theoretical yield of the aluminum. The number of moles produced theoretically can be used to calculate the amount of aluminum oxide required in moles and when we multiply the value obtained by the molar mass of aluminum oxide, we obtain the mass of aluminum oxide required.