Worksheet: Percent Yield Name Chlorobenzene, C6H6, is used in the ... - Free Printable

Let’s solve each part step by step.

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Problem 1:

Balanced equation:
C₆H₆ (l) + Cl₂ (g) → C₆H₅Cl (s) + HCl (g)

We’re told 45.6 g of benzene (C₆H₆) reacts.

Part a: Theoretical yield of chlorobenzene (C₆H₅Cl)

Step 1: Find molar mass of benzene (C₆H₆)
Carbon = 12.01 g/mol × 6 = 72.06
Hydrogen = 1.008 g/mol × 6 = 6.048
Total = 72.06 + 6.048 = 78.108 g/mol

Step 2: Moles of benzene used
moles = mass / molar mass = 45.6 g / 78.108 g/mol ≈ 0.5838 mol

Step 3: From the balanced equation, 1 mole of benzene produces 1 mole of chlorobenzene.
So, moles of chlorobenzene expected = 0.5838 mol

Step 4: Molar mass of chlorobenzene (C₆H₅Cl)
Carbon = 12.01 × 6 = 72.06
Hydrogen = 1.008 × 5 = 5.04
Chlorine = 35.45
Total = 72.06 + 5.04 + 35.45 = 112.55 g/mol

Step 5: Theoretical yield in grams
mass = moles × molar mass = 0.5838 mol × 112.55 g/mol ≈ 65.71 g

✔ So theoretical yield is about 65.7 g (rounded to one decimal place as per given data).

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Part b: Percent yield if actual yield is 63.7 g

Percent yield = (actual yield / theoretical yield) × 100%
= (63.7 / 65.71) × 100% ≈ 96.9%

✔ Percent yield ≈ 96.9%

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Problem 2:

Balanced equation:
CS₂ (l) + 3 O₂ (g) → CO₂ (g) + 2 SO₂ (g)

Part a: Percent yield of SO₂ when 25.0 g CS₂ produces 40.5 g SO₂

Step 1: Molar mass of CS₂
Carbon = 12.01
Sulfur = 32.07 × 2 = 64.14
Total = 12.01 + 64.14 = 76.15 g/mol

Step 2: Moles of CS₂ used
moles = 25.0 g / 76.15 g/mol ≈ 0.3283 mol

Step 3: From equation, 1 mol CS₂ produces 2 mol SO₂
→ moles of SO₂ expected = 0.3283 × 2 = 0.6566 mol

Step 4: Molar mass of SO₂
Sulfur = 32.07
Oxygen = 16.00 × 2 = 32.00
Total = 64.07 g/mol

Step 5: Theoretical yield of SO₂
mass = 0.6566 mol × 64.07 g/mol ≈ 42.07 g

Step 6: Percent yield
= (actual / theoretical) × 100% = (40.5 / 42.07) × 100% ≈ 96.3%

✔ Percent yield of SO₂ ≈ 96.3%

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Part b: Percent yield of CO₂ if 2.5 mol O₂ react and 32.4 g CO₂ are produced

Step 1: Use stoichiometry from balanced equation
3 mol O₂ → 1 mol CO₂
So, 2.5 mol O₂ will produce:
(1 mol CO₂ / 3 mol O₂) × 2.5 mol O₂ = 0.8333 mol CO₂

Step 2: Molar mass of CO₂
Carbon = 12.01
Oxygen = 16.00 × 2 = 32.00
Total = 44.01 g/mol

Step 3: Theoretical yield of CO₂
mass = 0.8333 mol × 44.01 g/mol ≈ 36.67 g

Step 4: Actual yield = 32.4 g
Percent yield = (32.4 / 36.67) × 100% ≈ 88.4%

✔ Percent yield of CO₂ ≈ 88.4%

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Final Answer:
1a. 65.7 g
1b. 96.9%
2a. 96.3%
2b. 88.4%