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To calculate the percent yield of carbon dioxide (\( \text{CO}_2 \)) produced in this reaction, let's go through the steps.

### Step 1: Write the Balanced Chemical Equation
The combustion of octane (\( \text{C}_8\text{H}_{18} \)) with oxygen gas (\( \text{O}_2 \)) is:

\[
\text{C}_8\text{H}_{18} + 12.5 \, \text{O}_2 \rightarrow 8 \, \text{CO}_2 + 9 \, \text{H}_2\text{O}
\]

Since we prefer whole numbers, multiply the entire equation by 2:

\[
2 \, \text{C}_8\text{H}_{18} + 25 \, \text{O}_2 \rightarrow 16 \, \text{CO}_2 + 18 \, \text{H}_2\text{O}
\]

This means 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide.

### Step 2: Calculate Molar Masses
- **Molar mass of octane (\( \text{C}_8\text{H}_{18} \))**:  
  \( (8 \times 12.01) + (18 \times 1.008) = 114.224 \, \text{g/mol} \)
  
- **Molar mass of oxygen (\( \text{O}_2 \))**:  
  \( 2 \times 16.00 = 32.00 \, \text{g/mol} \)

- **Molar mass of carbon dioxide (\( \text{CO}_2 \))**:  
  \( 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol} \)

### Step 3: Determine Limiting Reactant
We have:
- 78.81 g of octane
- 140.8 g of oxygen

Convert these masses to moles:

- **Moles of octane**:
  \[
  \frac{78.81 \, \text{g}}{114.224 \, \text{g/mol}} \approx 0.69 \, \text{moles}
  \]

- **Moles of oxygen**:
  \[
  \frac{140.8 \, \text{g}}{32.00 \, \text{g/mol}} = 4.40 \, \text{moles}
  \]

From the balanced equation, 2 moles of octane require 25 moles of oxygen. Therefore, 0.69 moles of octane would require:
\[
0.69 \times \frac{25}{2} = 8.625 \, \text{moles of } \text{O}_2
\]

Since we only have 4.40 moles of \( \text{O}_2 \), oxygen is the limiting reactant.

### Step 4: Calculate Theoretical Yield of \( \text{CO}_2 \)
Using the balanced equation, 25 moles of \( \text{O}_2 \) produce 16 moles of \( \text{CO}_2 \). Thus, 4.40 moles of \( \text{O}_2 \) will produce:
\[
4.40 \times \frac{16}{25} = 2.816 \, \text{moles of } \text{CO}_2
\]

Now, convert moles of \( \text{CO}_2 \) to grams:
\[
2.816 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 123.98 \, \text{g}
\]

So, the theoretical yield of \( \text{CO}_2 \) is approximately 123.98 g.

### Step 5: Calculate Percent Yield
The actual yield of \( \text{CO}_2 \) is given as 88.0 g. Now, calculate the percent yield:

\[
\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{88.0}{123.98} \times 100
\]

\[
\approx 70.98\%
\]

### Final Answer (Rounded to 3 Significant Figures)
The percent yield of \( \text{CO}_2 \) is **71.0%**.

Would you like more details on any step or have any questions?

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### Related Questions:
1. How do you determine the limiting reactant in a chemical reaction?
2. Why is it important to use the balanced chemical equation for stoichiometry calculations?
3. How do you calculate molar mass for a compound with multiple elements?
4. What are some common reasons for a reaction not reaching 100% yield?
5. How can percent yield be improved in a laboratory setting?

### Tip:
Always confirm the limiting reactant before calculating theoretical yield, as it directly affects the amount of product formed.

Liquid octane (C₈H₁₈) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). If 88.0 g of carbon dioxide is produced from the reaction of 78.81 g of octane and 140.8 g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.

Learn how to calculate the percent yield of carbon dioxide (CO₂) in a reaction involving octane (C₈H₁₈) and oxygen gas (O₂). This step-by-step solution covers concepts of stoichiometry, limiting reactants, and yield calculations. Suitable for high school chemistry students.

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