8.000 moles of Cl2 (g) react with excess P4 (s) to produce 0.5741 mol of phosphorus trichloride. What is the percent yield for this reaction? P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

%Yield ≈ 10.8% w/w

Explanation:

P₄ + 6Cl₂ => 4PCl₃

from equation rxn ratios 8 mol Cl₂ => 8(4/6) mole PCl₃ = 5.33 mole theoretically

Given Actual Yield = 0.5741 mole PCl₃;

%Yield = (Actual Yield / Theoretical Yield)100% = (0.5741/5.33)100% = 10.8%

The percent yield is 10.77 %

Explanation:

Step 1: Data given

Number of moles Cl2 = 8.00 moles

Number of moles PCl3 produced = 0.5741 moles

Step 2: The balanced equation

P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

Step 3: Calculate moles PCl3

For 1 mol P4 we need 6 moles Cl2 to produce 4 moles PCl3

For 8.00 moles Cl2 we'll have 4/6 * 8.00 = 5.33 moles

Step 4: Calculate percent yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (0.5741 / 5.33) *100%

% yield =10.77%

The percent yield is 10.77 %

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