Percent yield problems worksheet with chemical equations and mass data for practice.

Let’s solve each percent yield problem step by step. Percent yield is calculated as:

> % Yield = (Actual Yield / Theoretical Yield) × 100%

We’ll need to:
1. Balance the equation (already done).
2. Find molar masses.
3. Convert given masses to moles.
4. Determine limiting reactant (if needed).
5. Calculate theoretical yield of product.
6. Plug into formula.

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## 🔹 PROBLEM 1

Reaction:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Given:
- 32 g C₂H₆
- 39 g O₂
- Actual yield of CO₂ = 28 g
Find: % yield

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Step 1: Molar Masses

- C₂H₆: (2×12.01) + (6×1.008) = 24.02 + 6.048 = 30.068 g/mol
- O₂: 2×16.00 = 32.00 g/mol
- CO₂: 12.01 + (2×16.00) = 44.01 g/mol

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Step 2: Moles of Reactants

- Moles C₂H₆ = 32 g / 30.068 g/mol ≈ 1.064 mol
- Moles O₂ = 39 g / 32.00 g/mol ≈ 1.219 mol

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Step 3: Limiting Reactant

From balanced equation:
2 mol C₂H₆ : 7 mol O₂

So, for 1.064 mol C₂H₆, we need:

(7/2) × 1.064 = 3.724 mol O₂

But we only have 1.219 mol O₂ → O₂ is limiting reactant

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Step 4: Theoretical Yield of CO₂ from O₂

From equation:
7 mol O₂ → 4 mol CO₂

So, moles CO₂ = (4/7) × 1.219 ≈ 0.6966 mol

Mass CO₂ = 0.6966 mol × 44.01 g/mol ≈ 30.66 g

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Step 5: % Yield

% Yield = (Actual / Theoretical) × 100
= (28 / 30.66) × 100 ≈ 91.3%

✔ Answer: 91.3%

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## 🔹 PROBLEM 2

Reaction:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Given:
- 32 g C₂H₆ → produces 44 g CO₂ (actual)
Find: % yield

*(Note: O₂ amount not given — assume excess, so C₂H₆ is limiting)*

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Step 1: Molar Masses (same as above)

- C₂H₆ = 30.068 g/mol
- CO₂ = 44.01 g/mol

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Step 2: Moles C₂H₆

Moles = 32 g / 30.068 g/mol ≈ 1.064 mol

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Step 3: Theoretical Yield of CO₂

From equation:
2 mol C₂H₆ → 4 mol CO₂

So, moles CO₂ = (4/2) × 1.064 = 2.128 mol

Mass CO₂ = 2.128 × 44.01 ≈ 93.65 g

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Step 4: % Yield

% Yield = (44 / 93.65) × 100 ≈ 46.98%

✔ Answer: 47.0% (rounded to one decimal)

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## 🔹 PROBLEM 3

Reaction:
CS₂ + 3Cl₂ → CCl₄ + S₂Cl₂

Given:
- 80 g Cl₂ → produces 38 g CCl₄
Find: % yield

*(Assume CS₂ is in excess — Cl₂ is limiting)*

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Step 1: Molar Masses

- Cl₂ = 2×35.45 = 70.90 g/mol
- CCl₄ = 12.01 + (4×35.45) = 12.01 + 141.80 = 153.81 g/mol

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Step 2: Moles Cl₂

Moles = 80 g / 70.90 g/mol ≈ 1.128 mol

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Step 3: Theoretical Yield of CCl₄

From equation:
3 mol Cl₂ → 1 mol CCl₄

So, moles CCl₄ = (1/3) × 1.128 ≈ 0.376 mol

Mass CCl₄ = 0.376 × 153.81 ≈ 57.83 g

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Step 4: % Yield

% Yield = (38 / 57.83) × 100 ≈ 65.7%

✔ Answer: 65.7%

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## 🔹 PROBLEM 4

Reaction:
CH₃COF + H₂O → CH₃COOH + HF

Given:
- 4 g H₂O → produces 0.67 g HF
Find: % yield

*(Assume CH₃COF is in excess — H₂O is limiting)*

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Step 1: Molar Masses

- H₂O = (2×1.008) + 16.00 = 18.016 g/mol
- HF = 1.008 + 19.00 = 20.008 g/mol

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Step 2: Moles H₂O

Moles = 4 g / 18.016 g/mol ≈ 0.222 mol

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Step 3: Theoretical Yield of HF

From equation:
1 mol H₂O → 1 mol HF

So, moles HF = 0.222 mol

Mass HF = 0.222 × 20.008 ≈ 4.442 g

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Step 4: % Yield

% Yield = (0.67 / 4.442) × 100 ≈ 15.08%

✔ Answer: 15.1% (rounded to one decimal)

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## ✔ FINAL ANSWERS:

1. 91.3%
2. 47.0%
3. 65.7%
4. 15.1%

Let me know if you’d like to see the work with more precise atomic masses or step-by-step dimensional analysis!