Anhydrous aluminum oxide can be reduced to aluminum according to thisa chemical equation: 2 Al2O3 (s) ---> 4 Al (l) + 3 O2 (g) Calculate the mass of aluminum oxide reacted if 82 g of Al was produced and the percent yield was found to be 89.3%

Answer 1

This works the same way as the Cu(NO3)2 problem.
First, what mass Al would we have had at 100% yield; i.e., the theoretical yield? Thats
%yield = [(actual yield/theoretical yield)]*100
89.3% = (82/theor)*100. Solve for theoretical yield and use that below.
mols Al = grams/molar mass.
Using the coefficients, convert mols Al to mols Al2O3.
Then convert mols Al2O3 t grams.
I ran through and estimated 193g Al2O3 needed to produce 82g Al at 89.3% yield.
Be sure to confirm that.