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23 January, 11:12

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?

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  1. 23 January, 13:03
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    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 %
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