Asked by Sammy
when 4.50 g of fe2o3 is reduced with excess h2 in a furnace, 2.6 g of iron metal is recovered. What is the percent yield when molar mass of Fe2O3 is 159.7
Answers
Answered by
DrBob222
Fe2O3 + 3H2 ==> 3H2O + 2Fe
mols Fe2O3 = grams/molar mass
Convert mols Fe2O3 to mols Fe using the coefficients.
Convert mols Fe to grams. g = mols x molar mass. This is the theoretical yield.
%yield = (actual yield 2.6g/theore yield)*100 = ?
mols Fe2O3 = grams/molar mass
Convert mols Fe2O3 to mols Fe using the coefficients.
Convert mols Fe to grams. g = mols x molar mass. This is the theoretical yield.
%yield = (actual yield 2.6g/theore yield)*100 = ?
Answered by
Goon
(4.50g Fe2O3)*(1mol Fe2O3/159.7g Fe2O3)*(111.69g Fe2/1mol Fe2)= 3.15g Fe2
%Yield = (2.60g/3.15g)*100 = 0.825*100 = 82.5%
%Yield = (2.60g/3.15g)*100 = 0.825*100 = 82.5%
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