How to Solve: Reaction Yield

How to Solve: Reaction Yield

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"If you’ve ever wondered why your lab experiment didn’t produce as much product as the textbook said it should—this is the skill that explains it. And on your exam, it’s worth easy marks if you follow these steps."

What You Need To Know First

Before tackling reaction yield, you must understand: 1. Balanced chemical equations – How to write and interpret them (e.g., 2H₂ + O₂ → 2H₂O). 2. Mole concept – How to calculate moles from mass (moles = mass ÷ molar mass). 3. Limiting reactant – How to identify which reactant runs out first.

(If any of these are shaky, pause and review them first.)

Key Vocabulary

Formulas To Know

1. Percentage Yield

Formula: [ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% ] Variables: - Actual yield = Mass of product obtained in the lab (given or measured). - Theoretical yield = Maximum possible mass of product (calculated from the balanced equation).

MEMORISE THIS – It’s the most common yield question on exams.

2. Theoretical Yield (from moles)

Formula: [ \text{Theoretical yield (g)} = \text{Moles of limiting reactant} \times \text{Molar ratio} \times \text{Molar mass of product} ] Variables: - Moles of limiting reactant = Mass of limiting reactant ÷ its molar mass. - Molar ratio = Coefficients from the balanced equation (e.g., 2H₂O : 2H₂ → ratio = 1:1). - Molar mass of product = Sum of atomic masses (e.g., H₂O = 18 g/mol).

Given on exam sheet? Sometimes, but you must know how to use it.

Step-by-Step Method

(Follow these steps for every yield problem.)

1. Write the balanced equation.

2. Example: 2H₂ + O₂ → 2H₂O.

3. Calculate moles of each reactant.

4. Moles = mass (g) ÷ molar mass (g/mol).

5. Identify the limiting reactant.

6. Compare mole ratios to the balanced equation.

7. The reactant with fewer moles relative to its coefficient is limiting.

8. Calculate theoretical yield.

9. Use the limiting reactant’s moles × molar ratio × product’s molar mass.

10. Calculate percentage yield.

11. (Actual yield ÷ Theoretical yield) × 100%.

Worked Examples

Example 1 – Basic

Question: Hydrogen reacts with oxygen to form water. If 4g of H₂ reacts with excess O₂ and 30g of H₂O is produced, what is the percentage yield?

Solution: 1. Balanced equation: 2H₂ + O₂ → 2H₂O. 2. Moles of H₂: 4g ÷ 2 g/mol = 2 mol. 3. Limiting reactant: H₂ (O₂ is in excess). 4. Theoretical yield:
- Molar ratio (H₂O : H₂) = 2:2 → 1:1.
- Moles of H₂O = 2 mol × 1 = 2 mol.
- Mass of H₂O = 2 mol × 18 g/mol = 36g. 5. Percentage yield: (30g ÷ 36g) × 100% = 83.3%.

What we did and why: - We used the limiting reactant (H₂) to find the maximum possible product (36g). - The actual yield (30g) was less, so we calculated how efficient the reaction was (83.3%).

Example 2 – Medium

Question: 10g of calcium carbonate (CaCO₃) decomposes to form calcium oxide (CaO) and CO₂. If 4.5g of CaO is obtained, what is the percentage yield?

Solution: 1. Balanced equation: CaCO₃ → CaO + CO₂. 2. Moles of CaCO₃: 10g ÷ 100 g/mol = 0.1 mol. 3. Limiting reactant: CaCO₃ (only reactant). 4. Theoretical yield:
- Molar ratio (CaO : CaCO₃) = 1:1.
- Moles of CaO = 0.1 mol × 1 = 0.1 mol.
- Mass of CaO = 0.1 mol × 56 g/mol = 5.6g. 5. Percentage yield: (4.5g ÷ 5.6g) × 100% = 80.4%.

What we did and why: - We assumed 100% decomposition to find the theoretical yield (5.6g). - The actual yield (4.5g) was lower, so we calculated the efficiency (80.4%).

Example 3 – Exam Style

Question: In the reaction 2Al + 3Cl₂ → 2AlCl₃, 5.4g of Al reacts with 14.2g of Cl₂. If 13.35g of AlCl₃ is produced, what is the percentage yield?

Solution: 1. Balanced equation: 2Al + 3Cl₂ → 2AlCl₃. 2. Moles of Al: 5.4g ÷ 27 g/mol = 0.2 mol.
Moles of Cl₂: 14.2g ÷ 71 g/mol = 0.2 mol. 3. Limiting reactant:
- Required ratio (Al : Cl₂) = 2:3.
- Available ratio = 0.2:0.2 = 1:1.
- Cl₂ is limiting (needs 0.3 mol for 0.2 mol Al, but only 0.2 mol available). 4. Theoretical yield:
- Molar ratio (AlCl₃ : Cl₂) = 2:3.
- Moles of AlCl₃ = 0.2 mol Cl₂ × (2/3) = 0.133 mol.
- Mass of AlCl₃ = 0.133 mol × 133.5 g/mol = 17.8g. 5. Percentage yield: (13.35g ÷ 17.8g) × 100% = 75%.

What we did and why: - We identified Cl₂ as the limiting reactant by comparing mole ratios. - The theoretical yield (17.8g) was higher than the actual yield (13.35g), giving 75% efficiency.

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in. Reaction yield is all about comparing what you should get (theoretical yield) to what you actually get (actual yield). Here’s the drill: 1. Balance the equation – No shortcuts here. 2. Find moles of each reactant – Mass ÷ molar mass. 3. Pick the limiting reactant – The one that runs out first. 4. Calculate theoretical yield – Use the limiting reactant’s moles × molar ratio × product’s molar mass. 5. Plug into percentage yield – (Actual ÷ Theoretical) × 100%.

Watch out for traps: excess reactants, hidden units, and molar ratio mix-ups. Do one step at a time, and you’ll nail it. Now go crush that exam!