If the percent yield was 81.5%, how many grams of aluminum were used to prepare 20.0 g of
potassium alum?

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  1. The skeleton equation is
    Al ==> KAl(SO4)2.12H2O

    20g final product. 81.5% yield means
    20/0.815 = 24.5 g KAl(SO4)2.12H2O formed.
    mol = 24.5/molar mass KAl(SO4)2.12H2O
    That many mol Al were required.
    g Al = mols x atomic mass Al.

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