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Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. (Molar masses: Fe = 55.85 g/mol, O = 16.00 g/mol, Fe2O3 = 159.70 g/mol)
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
A) 61.03 %
B) 28.65 %
C) 57.30 %
D) 20.02 %
E) 81.93 %
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