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Problem Analysis:

The image contains a chemistry worksheet focusing on percent yield calculations. Percent yield is a measure of the efficiency of a chemical reaction, calculated using the formula:

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

We will solve each part of the problems step by step.

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Problem 1: Chlorobenzene Production

#### Given:
- Balanced equation:
\[
C_6H_6(l) + Cl_2(g) \rightarrow C_6H_5Cl(s) + HCl(g)
\]
- Mass of benzene (\(C_6H_6\)) = 45.6 g
- Actual yield of chlorobenzene (\(C_6H_5Cl\)) = 63.7 g

#### Part (a): Theoretical Yield of Chlorobenzene

1. Molar Mass Calculation:
- Molar mass of \(C_6H_6\):
\[
6 \times 12.01 \, \text{(C)} + 6 \times 1.01 \, \text{(H)} = 78.12 \, \text{g/mol}
\]
- Molar mass of \(C_6H_5Cl\):
\[
6 \times 12.01 \, \text{(C)} + 5 \times 1.01 \, \text{(H)} + 35.45 \, \text{(Cl)} = 112.56 \, \text{g/mol}
\]

2. Moles of Benzene:
\[
\text{Moles of } C_6H_6 = \frac{\text{Mass of } C_6H_6}{\text{Molar Mass of } C_6H_6} = \frac{45.6 \, \text{g}}{78.12 \, \text{g/mol}} \approx 0.584 \, \text{mol}
\]

3. Stoichiometry:
From the balanced equation, 1 mole of \(C_6H_6\) produces 1 mole of \(C_6H_5Cl\). Therefore:
\[
\text{Moles of } C_6H_5Cl = \text{Moles of } C_6H_6 = 0.584 \, \text{mol}
\]

4. Theoretical Yield of Chlorobenzene:
\[
\text{Theoretical Yield} = \text{Moles of } C_6H_5Cl \times \text{Molar Mass of } C_6H_5Cl
\]
\[
\text{Theoretical Yield} = 0.584 \, \text{mol} \times 112.56 \, \text{g/mol} \approx 65.7 \, \text{g}
\]

Answer for Part (a):
\[
\boxed{65.7 \, \text{g}}
\]

#### Part (b): Percent Yield

1. Formula for Percent Yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]

2. Substitute Values:
\[
\text{Percent Yield} = \left( \frac{63.7 \, \text{g}}{65.7 \, \text{g}} \right) \times 100 \approx 96.9\%
\]

Answer for Part (b):
\[
\boxed{96.9\%}
\]

---

Problem 2: Carbon Disulfide Combustion

#### Given:
- Balanced equation:
\[
CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)
\]
- Mass of carbon disulfide (\(CS_2\)) = 25.0 g
- Actual yield of sulfur dioxide (\(SO_2\)) = 40.5 g
- Moles of oxygen (\(O_2\)) = 2.5 mol
- Actual yield of carbon dioxide (\(CO_2\)) = 32.4 g

#### Part (a): Percent Yield of Sulfur Dioxide (\(SO_2\))

1. Molar Mass Calculation:
- Molar mass of \(CS_2\):
\[
12.01 \, \text{(C)} + 2 \times 32.07 \, \text{(S)} = 76.15 \, \text{g/mol}
\]
- Molar mass of \(SO_2\):
\[
32.07 \, \text{(S)} + 2 \times 16.00 \, \text{(O)} = 64.07 \, \text{g/mol}
\]

2. Moles of Carbon Disulfide:
\[
\text{Moles of } CS_2 = \frac{\text{Mass of } CS_2}{\text{Molar Mass of } CS_2} = \frac{25.0 \, \text{g}}{76.15 \, \text{g/mol}} \approx 0.328 \, \text{mol}
\]

3. Stoichiometry:
From the balanced equation, 1 mole of \(CS_2\) produces 2 moles of \(SO_2\). Therefore:
\[
\text{Moles of } SO_2 = 2 \times \text{Moles of } CS_2 = 2 \times 0.328 \, \text{mol} \approx 0.656 \, \text{mol}
\]

4. Theoretical Yield of Sulfur Dioxide:
\[
\text{Theoretical Yield} = \text{Moles of } SO_2 \times \text{Molar Mass of } SO_2
\]
\[
\text{Theoretical Yield} = 0.656 \, \text{mol} \times 64.07 \, \text{g/mol} \approx 42.0 \, \text{g}
\]

5. Percent Yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
\[
\text{Percent Yield} = \left( \frac{40.5 \, \text{g}}{42.0 \, \text{g}} \right) \times 100 \approx 96.4\%
\]

Answer for Part (a):
\[
\boxed{96.4\%}
\]

#### Part (b): Percent Yield of Carbon Dioxide (\(CO_2\))

1. Molar Mass Calculation:
- Molar mass of \(CO_2\):
\[
12.01 \, \text{(C)} + 2 \times 16.00 \, \text{(O)} = 44.01 \, \text{g/mol}
\]

2. Moles of Oxygen:
\[
\text{Moles of } O_2 = 2.5 \, \text{mol}
\]

3. Stoichiometry:
From the balanced equation, 3 moles of \(O_2\) produce 1 mole of \(CO_2\). Therefore:
\[
\text{Moles of } CO_2 = \frac{1}{3} \times \text{Moles of } O_2 = \frac{1}{3} \times 2.5 \, \text{mol} \approx 0.833 \, \text{mol}
\]

4. Theoretical Yield of Carbon Dioxide:
\[
\text{Theoretical Yield} = \text{Moles of } CO_2 \times \text{Molar Mass of } CO_2
\]
\[
\text{Theoretical Yield} = 0.833 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 36.7 \, \text{g}
\]

5. Percent Yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
\[
\text{Percent Yield} = \left( \frac{32.4 \, \text{g}}{36.7 \, \text{g}} \right) \times 100 \approx 88.3\%
\]

Answer for Part (b):
\[
\boxed{88.3\%}
\]

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Final Answers:

1. (a) Theoretical yield of chlorobenzene: \(\boxed{65.7 \, \text{g}}\)
(b) Percent yield of chlorobenzene: \(\boxed{96.9\%}\)

2. (a) Percent yield of sulfur dioxide: \(\boxed{96.4\%}\)
(b) Percent yield of carbon dioxide: \(\boxed{88.3\%}\)