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Problem: 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)  →  Al 2O3 (s) + 2 Fe (s)A. 61.03 %B. 28.65 %C. 57.30 %D. 20.02 %E. 81.93 %

🤓 Based on our data, we think this question is relevant for Professor Edwards III's class at FAMU.

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Problem Details

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)  →  Al 2O(s) + 2 Fe (s)

A. 61.03 %

B. 28.65 %

C. 57.30 %

D. 20.02 %

E. 81.93 %

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