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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
- Molar mass of \(C_6H_6\) = 78.11 g/mol
- Molar mass of \(C_6H_5Cl\) = 112.56 g/mol
- Actual yield of chlorobenzene (\(C_6H_5Cl\)) = 63.7 g

#### Part (a): Theoretical Yield

1. Calculate the moles of benzene (\(C_6H_6\)):
\[
\text{Moles of } C_6H_6 = \frac{\text{Mass of } C_6H_6}{\text{Molar mass of } C_6H_6} = \frac{45.6 \, \text{g}}{78.11 \, \text{g/mol}} \approx 0.584 \, \text{mol}
\]

2. Determine the moles of chlorobenzene (\(C_6H_5Cl\)) produced:
From the balanced equation, 1 mole of \(C_6H_6\) produces 1 mole of \(C_6H_5Cl\). Therefore:
\[
\text{Moles of } C_6H_5Cl = \text{Moles of } C_6H_6 = 0.584 \, \text{mol}
\]

3. Calculate the theoretical yield of chlorobenzene in grams:
\[
\text{Theoretical yield of } C_6H_5Cl = \text{Moles of } C_6H_5Cl \times \text{Molar mass of } C_6H_5Cl
\]
\[
\text{Theoretical yield of } C_6H_5Cl = 0.584 \, \text{mol} \times 112.56 \, \text{g/mol} \approx 65.7 \, \text{g}
\]

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

#### Part (b): Percent Yield

1. Use the formula for percent yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]

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

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

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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
- Molar mass of \(CS_2\) = 76.14 g/mol
- Molar mass of \(SO_2\) = 64.07 g/mol
- Molar mass of \(CO_2\) = 44.01 g/mol
- Actual yield of \(SO_2\) = 40.5 g
- Moles of oxygen (\(O_2\)) = 2.5 mol
- Actual yield of \(CO_2\) = 32.4 g

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

1. Calculate the moles of \(CS_2\):
\[
\text{Moles of } CS_2 = \frac{\text{Mass of } CS_2}{\text{Molar mass of } CS_2} = \frac{25.0 \, \text{g}}{76.14 \, \text{g/mol}} \approx 0.328 \, \text{mol}
\]

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

3. Calculate the theoretical yield of \(SO_2\) in grams:
\[
\text{Theoretical yield of } SO_2 = \text{Moles of } SO_2 \times \text{Molar mass of } SO_2
\]
\[
\text{Theoretical yield of } SO_2 = 0.656 \, \text{mol} \times 64.07 \, \text{g/mol} \approx 42.0 \, \text{g}
\]

4. Calculate the percent yield of \(SO_2\):
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
\[
\text{Percent Yield} = \left( \frac{40.5 \, \text{g}}{42.0 \, \text{g}} \right) \times 100 \approx 96.4\%
\]

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

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

1. Determine the limiting reactant:
- Moles of \(CS_2\): Already calculated as 0.328 mol.
- Moles of \(O_2\): Given as 2.5 mol.
- From the balanced equation, 1 mole of \(CS_2\) requires 3 moles of \(O_2\). Therefore, 0.328 mol of \(CS_2\) requires:
\[
\text{Moles of } O_2 \text{ required} = 3 \times 0.328 \, \text{mol} \approx 0.984 \, \text{mol}
\]
- Since 2.5 mol of \(O_2\) is available, \(O_2\) is in excess, and \(CS_2\) is the limiting reactant.

2. Calculate the moles of \(CO_2\) produced:
From the balanced equation, 1 mole of \(CS_2\) produces 1 mole of \(CO_2\). Therefore:
\[
\text{Moles of } CO_2 = \text{Moles of } CS_2 = 0.328 \, \text{mol}
\]

3. Calculate the theoretical yield of \(CO_2\) in grams:
\[
\text{Theoretical yield of } CO_2 = \text{Moles of } CO_2 \times \text{Molar mass of } CO_2
\]
\[
\text{Theoretical yield of } CO_2 = 0.328 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 14.4 \, \text{g}
\]

4. Calculate the percent yield of \(CO_2\):
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
\[
\text{Percent Yield} = \left( \frac{32.4 \, \text{g}}{14.4 \, \text{g}} \right) \times 100 \approx 225\%
\]

Answer for Part (b):
\[
\boxed{225\%}
\]

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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: \(\boxed{225\%}\)