The balanced equation for the combustion of butane is: 2C4H10+1302 8CO₂ +10H₂0 di See Periodic Table % See F If the combustion of 48.65 g of C4H10 produces 108.55 g of CO₂. What is the percent yield of the reaction? (Assume oxygen is in excess.)
The balanced equation for the combustion of butane is: 2C4H10+1302 8CO₂ +10H₂0 di See Periodic Table % See F If the combustion of 48.65 g of C4H10 produces 108.55 g of CO₂. What is the percent yield of the reaction? (Assume oxygen is in excess.)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Combustion of Butane**
The balanced equation for the combustion of butane is:
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
---
To access additional resources:
- [See Periodic Table](#)
- [See Hint](#)
---
**Problem Statement**
If the combustion of 48.65 g of \( C_4H_{10} \) produces 108.55 g of \( CO_2 \), what is the percent yield of the reaction? (Assume oxygen is in excess.)
**Enter your answer:**
\[ \_\_\_\_ \% \]
---
**Explanation for Percent Yield Calculation**
Percent yield is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
1. **Find Moles of \( C_4H_{10} \):**
\[ \text{Molar Mass of } C_4H_{10} = 4 \times 12.01 \, \text{g/mol} + 10 \times 1.008 \, \text{g/mol} = 58.14 \, \text{g/mol} \]
\[ \text{Moles of } C_4H_{10} = \frac{48.65 \, \text{g}}{58.14 \, \text{g/mol}} \approx 0.837 \, \text{mol} \]
2. **Calculate Theoretical Yield of \( CO_2 \):**
- From the balanced equation: \( 2 \text{mol of } C_4H_{10} \rightarrow 8 \text{mol of } CO_2 \)
- Therefore: \( 0.837 \text{mol of } C_4H_{10} \rightarrow \frac{8}{2} \times 0.837 \text{mol of } CO_2 \approx 3.348 \text{mol of } CO_2 \)
\[ \text{Molar Mass of } CO_2 = 44.01 \, \text{g/mol} \]
\[ \text{Theoretical Yield} = 3.348 \, \text{mol} \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c2f3fb5-efee-4c32-9033-75e1b446c804%2F83b1bd8b-4c1d-4d9a-b60f-7b35aabff431%2Fafvz7yg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Combustion of Butane**
The balanced equation for the combustion of butane is:
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
---
To access additional resources:
- [See Periodic Table](#)
- [See Hint](#)
---
**Problem Statement**
If the combustion of 48.65 g of \( C_4H_{10} \) produces 108.55 g of \( CO_2 \), what is the percent yield of the reaction? (Assume oxygen is in excess.)
**Enter your answer:**
\[ \_\_\_\_ \% \]
---
**Explanation for Percent Yield Calculation**
Percent yield is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
1. **Find Moles of \( C_4H_{10} \):**
\[ \text{Molar Mass of } C_4H_{10} = 4 \times 12.01 \, \text{g/mol} + 10 \times 1.008 \, \text{g/mol} = 58.14 \, \text{g/mol} \]
\[ \text{Moles of } C_4H_{10} = \frac{48.65 \, \text{g}}{58.14 \, \text{g/mol}} \approx 0.837 \, \text{mol} \]
2. **Calculate Theoretical Yield of \( CO_2 \):**
- From the balanced equation: \( 2 \text{mol of } C_4H_{10} \rightarrow 8 \text{mol of } CO_2 \)
- Therefore: \( 0.837 \text{mol of } C_4H_{10} \rightarrow \frac{8}{2} \times 0.837 \text{mol of } CO_2 \approx 3.348 \text{mol of } CO_2 \)
\[ \text{Molar Mass of } CO_2 = 44.01 \, \text{g/mol} \]
\[ \text{Theoretical Yield} = 3.348 \, \text{mol} \times
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