Question

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

Answer 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.

Answer 2

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