posted by Kelsey .
Please be specific. In an experiement 23.4 g of iron sulphide, FeS are added to excess oxygen, and 16.5 of Iron oxide, Fe2O3 are produced. The balanced euation for the reaction is 4FeS =7O2 -- 2Fe2O +2SO2. Calculate the percent yield of iron oxide in the experiment.
Step 1. Calculate the theoretical yield.
Step 2. Calculate percent yield.
One reason you may be having trouble is that the equation you have is not balanced.
You need a little work on the equation.
4FeS + 7O2 --> 2Fe2O3 + 4SO2
I will estimate the molar masses; you should do it more exacting than I but this will show you the procedure.
Convert 23.4 g FeS to moles. moles = grams/molar mass
23.4/88 = 0.266
Using the coefficients in the balanced equation, convert moles FeS to moles Fe2O3.
0.266 moles FeS x (2 moles Fe2O3/4 moles FeS) = 0.266 x (2/4) = 0.266 x (1/2) = 0.133 moles Fe2O3.
Now convert moles Fe2O3 to grams. g = moles x molar mass
0.133 x 160 = 21.3 g Fe2O3. This is the theoretical yield
%yield = (actual/theoretical)*100 =
(16.5/21.3)*100 = ??%
Check my work.