Chemistry worksheet on percent yield with detailed solutions for four different chemical reactions.

The image you've uploaded contains a chemistry worksheet on percent yield, along with the key (solutions) for four problems. Below is a detailed explanation of each problem and its solution, as shown in the key.

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🔬 Chemistry: Percent Yield

Percent yield is calculated using the formula:

$$
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
$$

We use stoichiometry to find the theoretical yield based on the balanced chemical equation, then compare it to the actual yield from the experiment.

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## ✔ Problem 1: Slaked Lime Production

Reaction:
$$
\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2
$$

- Start with: 2400 g CaO
- Actual yield of Ca(OH)₂: 2060 g
- Water is in excess → CaO is the limiting reactant

Step-by-step Solution:

1. Find moles of CaO:
$$
\frac{2400\ \text{g CaO}}{56\ \text{g/mol}} = 42.857\ \text{mol CaO}
$$

2. Use stoichiometry to find theoretical yield of Ca(OH)₂:
- 1 mol CaO → 1 mol Ca(OH)₂
- So, 42.857 mol CaO → 42.857 mol Ca(OH)₂

3. Convert to grams:
$$
42.857\ \text{mol} \times 74\ \text{g/mol} = 3171\ \text{g Ca(OH)}_2
$$

4. Calculate percent yield:
$$
\% \text{Yield} = \left( \frac{2060}{3171} \right) \times 100\% = 65\%
$$

✔ Answer: 65%

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## ✔ Problem 2: Thermite Reaction

Reaction:
$$
\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}
$$

- Aluminum mass: 258 g
- Rust (Fe₂O₃) is in excess → Al is limiting reactant
- Actual yield of Fe: 464 g

Step-by-step Solution:

1. Moles of Al:
$$
\frac{258\ \text{g}}{27\ \text{g/mol}} = 9.556\ \text{mol Al}
$$

2. Stoichiometry:
- 2 mol Al → 2 mol Fe → 1:1 ratio
- So, 9.556 mol Al → 9.556 mol Fe

3. Mass of Fe (theoretical):
$$
9.556\ \text{mol} \times 55.8\ \text{g/mol} = 533\ \text{g Fe}
$$

4. Percent yield:
$$
\left( \frac{464}{533} \right) \times 100\% = 87\%
$$

✔ Answer: 87%

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## ✔ Problem 3: Sulfur Dioxide from Zinc Sulfide

Reaction:
$$
2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2
$$

- Molecules of ZnS: $1.5 \times 10^{27}$
- Oxygen is in excess → ZnS is limiting
- Percent yield = 75%
- Find actual volume of SO₂ at STP

Step-by-step Solution:

1. Convert molecules to moles:
$$
\frac{1.5 \times 10^{27}\ \text{molecules}}{6.02 \times 10^{23}\ \text{molecules/mol}} = 2.49 \times 10^3\ \text{mol ZnS}
$$

2. Stoichiometry:
- 2 mol ZnS → 2 mol SO₂ → 1:1 ratio
- So, $2.49 \times 10^3$ mol ZnS → $2.49 \times 10^3$ mol SO₂

3. Volume at STP (1 mol = 22.4 L):
$$
2.49 \times 10^3\ \text{mol} \times 22.4\ \text{L/mol} = 5.58 \times 10^4\ \text{L SO}_2
$$

→ This is the theoretical yield.

4. Actual yield (75% of theoretical):
$$
0.75 \times 5.58 \times 10^4 = 4.19 \times 10^4\ \text{L SO}_2
$$

✔ Answer: $4.19 \times 10^4\ \text{L SO}_2$

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## ✔ Problem 4: Haber Process

Reaction:
$$
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
$$

- Desired actual yield of NH₃: 700 g
- Percent yield: 70%
- Hydrogen is in excess → N₂ is limiting
- Find mass of N₂ needed

Step-by-step Solution:

1. Find theoretical yield of NH₃:
$$
\text{Percent Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100 \\
70 = \frac{700}{\text{Theoretical}} \times 100 \\
\Rightarrow \text{Theoretical} = \frac{700}{0.70} = 1000\ \text{g NH}_3
$$

2. Moles of NH₃ (theoretical):
$$
\frac{1000\ \text{g}}{17\ \text{g/mol}} = 58.82\ \text{mol NH}_3
$$

3. Stoichiometry:
- 2 mol NH₃ ← 1 mol N₂
- So, moles of N₂ needed = $ \frac{58.82}{2} = 29.41\ \text{mol N}_2 $

4. Mass of N₂:
$$
29.41\ \text{mol} \times 28\ \text{g/mol} = 824\ \text{g N}_2
$$

✔ Answer: 824 g N₂

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📝 Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 65% |
| 2 | 87% |
| 3 | $4.19 \times 10^4\ \text{L SO}_2$ |
| 4 | 824 g N₂ |

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💡 Key Concepts Recap:

- Theoretical yield: Maximum amount possible based on stoichiometry.
- Actual yield: What you actually get in the lab.
- Percent yield: Measures efficiency of reaction.
- Always identify the limiting reactant.
- Use molar masses, balanced equations, and unit conversions carefully.

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