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.

Answer 1

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.