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If the reaction of 2.5 g of Al with 2.5 g of O2 produced 3.5 g of Al2O3, what is the percent yield of Al2)3? Be sure to write and balance equation.

4Al + 302 ---> 2Al2O3

2.5 g Al x (1 mol Al / 108 g) x (2 mol Al2O3 / 4 mol Al) = .0115 mol Al2O3

.0115 mol Al2O3 (204 g Al2O3 / 1 mol Al2O3) = 2.36 Al2O3

Percent Yield = (2.36 / 3.50) X 100 = 67 %

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