Determine the percent yield of a reaction that produces 28 65 g of Fe when 50.0 00 g of Fez0s react with excess Al according to the following reaction Fe2O3(s)+2 Al(s) Al2O3(s)+2 Fe(s)
The answer is a ( 81.9%).
moles of Fe2O3 = weight in gram / molecular mass = 50 / 159.69 = 0.313 moles
According to the given reaction ; 1 mole of Fe2O3 produce 2 moles of Fe
so, 0.313 moles of fe2O3 will produce = 0.313 x 2 = 0.626 moles of Fe
weight of Fe in grams:
moles = weight in gram / molecular mass
weight in grams = 0.626 x 55.847 = 34.97 gram Fe (theoritical yield)
% yield = (Actual yield / theoritical yield ) x 100
% yield = (28.65 / 34.97) x 100 = 0.819 x 100 = 81.9 %
Determine the percent yield of a reaction that produces 28 65 g of Fe when 50.0...
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