Answer:
FeCl3 is the limiting reactant
O2 is in excess
Theoretical yield Cl2 = 9.84 grams
The % yield is 96.5 %
Explanation:
Step 1: Data given
Mass of FeCl3 = 15.0 grams
Moles O2 = 4.0 moles
Mass of Cl2 produced = 9.5 grams
Step 2: The balanced equation
4FeCl3 + 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 limiting reactant
FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2
There will remain 4.0 - 0.069375 = 3.930625 moles
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 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.
Step 6: Calculate mass Cl2
Mass Cl2 = moles * molar mass
Mass Cl2 = 0.13875 moles * 70.9 g/mol
Mass Cl2 = 9.84 grams
Step 7: Calculate % yield
% yield = (actual yield / theoretical yield) * 100%
% yield = (9.5 grams / 9.84 grams ) * 100%
% yield = 96.5 %
The % yield is 96.5 %
