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Theoretical Yield of Alum?

A student reported an actual yield of 2.150 g of alum from the reaction of 0.465 g of aluminum metal. what is the theoretical & % yield

  • Answer:

    Use 2.5g Al as a reactant. Use 3.5g Al2O3 as a product. Using stoichiometry you need to convert grams of reactant to grams of product. So. 2.5gAl (1mol/27.0)(2molAl2O3/4molAl)(102g/mol/1… So. (102/27) x (2/4) x (2.5) = 4.72g. So now you divide the product mass over your new value (the maximum yield). 3.50/4.72 = .742. x 100 = 74.2 % yield.

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Use 2.5g Al as a reactant. Use 3.5g Al2O3 as a product. Using stoichiometry you need to convert grams of reactant to grams of product. So. 2.5gAl (1mol/27.0)(2molAl2O3/4molAl)(102g/mol/1… So. (102/27) x (2/4) x (2.5) = 4.72g. So now you divide the product mass over your new value (the maximum yield). 3.50/4.72 = .742. x 100 = 74.2 % yield.

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