How many grams of chlorine gas can be produced if 15 grams of FeCl3 reacts with 4 moles of O2? What is the limiting reactant? What is the excess reactant? If 9.5 grams of Cl2 gas is produced, what is the percent yield?
4 FeCl3 + 3O2 --> 2Fe2O3 + 6Cl2
In paragraph form write out the steps used to solve the above stoichiometry calculation. Explain how do we find the limiting reactant, excess reactant, theoretical yield, actual yield and calculate the percent yield?

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 %

im pretty sure its b