Calculate the theoretical yield of iron(III) oxide (Fe2O3) in grams.
The reaction is 2 Fe + O2 ------- 2 FeO
4.8 g O2 = 4.8/32 = 0.15 mol.
0.15 mol O2 produces 2 x 0.15 = 0.3 mol FeO. Multiplying this by the molar mass of FeO (72) produces 21.6 grams.
Theoretical yield = 21.6 grams
% yield = 18.4/21.6 = 85%
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