Answer:
The limiting reactant is FeCl3
The excess reactant is O2
The theoretical yield Cl2 is 9.84 grams
The % yield = 96.5 %
Explanation:
Step 1: Data given
Mass of FeCl3 = 15.0 grams
Moles of O2 = 4.0 moles
Mass of Cl2 = 9.5 grams = actual yield
Step 2: The balanced equation
4 FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
Step 3: Calculate moles FeCl3
Moles FeCl3 = mass FeCl3 / molar mass FeCl3
Moles FeCl3 = 15.0 grams / 162.2 g/mol
Moles FeCl3 = 0.0925 moles
Step 4: Calculate the limiting reactant
For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2
FeCl3 has the smallest amount of moles, this is the limiting reactant. It will be completely consumed ( 0.0925 moles).
O2 is in excess. There will react 3/4 * 0.0925 = 0.0694 moles
There will remain 4.0 - 0.0694 = 3.3904 moles O2
Step 5: Calculate moles Cl2
For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2
For 0.0925 moles FeCl3 we'll have 6/4 * 0.0925 = 0.13875 moles Cl2
Step 6: Calculate mass of Cl2
Mass Cl2 = moles Cl2 * molar mass Cl2
Mass Cl2 = 0.13875 moles * 70.9 g/mol
Mass Cl2 = 9.84 grams = theoretical yield
Step 7: Calculate % yield
% yield = (actual yield / theoretical yield) *100%
% yield = (9.5 grams / 9.84 grams) * 100%
% yield = 96.5 %
