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.
To calculate the percent yield of iron oxide in the experiment, we need to compare the actual yield of Fe2O3 (iron oxide) with the theoretical yield. The percent yield formula is given by:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
1. First, we need to find the theoretical yield of Fe2O3. We can determine this by stoichiometry, using the balanced equation provided:
FeS + 7/2 O2 -> Fe2O3 + SO2
From the equation, we see that 4 moles of FeS produce 2 moles of Fe2O3. Therefore, we can set up a ratio using the molar masses:
(4 moles FeS / 87.91 g) = (2 moles Fe2O3 / X g)
Solving for X, we find that the molar mass of Fe2O3 is 159.7 g/mol.
2. Now we can calculate the theoretical yield of Fe2O3. Given the mass of FeS (23.4 g), we can use stoichiometry and molar masses to find the corresponding mass of Fe2O3:
(23.4 g FeS) x (159.7 g Fe2O3 / 87.91 g FeS) = 42.5 g Fe2O3 (rounded to one decimal place)
Therefore, the theoretical yield of Fe2O3 is 42.5 g.
3. Next, we need to determine the actual yield of Fe2O3, which is given as 16.5 g.
4. Finally, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100 = (16.5 g / 42.5 g) × 100 = 38.82% (rounded to two decimal places)
Therefore, the percent yield of iron oxide in the experiment is approximately 38.82%.