Balance the chemical equation for the combustion of pentane: C5H12 (g) + O2(g) → CO2 (g) + H2O (g) I need help with the amount on reactant side, element, amount on product side
Balance the chemical equation for the combustion of pentane: C5H12 (g) + O2(g) → CO2 (g) + H2O (g) I need help with the amount on reactant side, element, amount on product side
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
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Balance the chemical equation for the combustion of pentane: C5H12 (g) + O2(g) → CO2 (g) + H2O (g)
I need help with the amount on reactant side, element, amount on product side
![**Balancing the Chemical Equation for the Combustion of Pentane**
**Equation to Balance:**
\[ \text{C}_5\text{H}_{12}(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \]
**Balanced Equation:**
\[ \text{C}_5\text{H}_{12}(g) + 8 \text{O}_2(g) \rightarrow 5 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \]
**Steps:**
1. **Carbon Atoms:**
- There are 5 carbon atoms in C\(_5\)H\(_{12}\).
- Multiply CO\(_2\) by 5: \(1 \times 5 = 5\).
2. **Hydrogen Atoms:**
- There are 12 hydrogen atoms in C\(_5\)H\(_{12}\).
- Multiply H\(_2\)O by 6: \(2 \times 6 = 12\).
3. **Oxygen Atoms:**
- Oxygen in CO\(_2\): 5 molecules \(\times\) 2 = 10 atoms.
- Oxygen in H\(_2\)O: 6 molecules \(\times\) 1 = 6 atoms.
- Total = 10 + 6 = 16 atoms.
- Divide total oxygen atoms by 2 to find the number of O\(_2\) molecules: \( \frac{16}{2} = 8 \).
**Summary Chart:**
| Amount Reactant Side | Element | Amount on Product Side |
|----------------------|---------|-------------------------|
| 5 | C | 5 |
| 12 | H | 12 |
| 16 | O | 16 |
**Status:**
- The equation is balanced, as indicated by the check mark and the word "balanced" written next to the table.
**Navigation:**
- Arrows indicate options to "Go back," "Click here for a hint," "Click here to check your answer," or "Go to next question."](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67fb381f-b1d1-4d61-a9f9-5797b47c778f%2Fe0de7a85-4623-4cbf-a58e-4a7b54e07f21%2F7s3k3nn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Balancing the Chemical Equation for the Combustion of Pentane**
**Equation to Balance:**
\[ \text{C}_5\text{H}_{12}(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \]
**Balanced Equation:**
\[ \text{C}_5\text{H}_{12}(g) + 8 \text{O}_2(g) \rightarrow 5 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \]
**Steps:**
1. **Carbon Atoms:**
- There are 5 carbon atoms in C\(_5\)H\(_{12}\).
- Multiply CO\(_2\) by 5: \(1 \times 5 = 5\).
2. **Hydrogen Atoms:**
- There are 12 hydrogen atoms in C\(_5\)H\(_{12}\).
- Multiply H\(_2\)O by 6: \(2 \times 6 = 12\).
3. **Oxygen Atoms:**
- Oxygen in CO\(_2\): 5 molecules \(\times\) 2 = 10 atoms.
- Oxygen in H\(_2\)O: 6 molecules \(\times\) 1 = 6 atoms.
- Total = 10 + 6 = 16 atoms.
- Divide total oxygen atoms by 2 to find the number of O\(_2\) molecules: \( \frac{16}{2} = 8 \).
**Summary Chart:**
| Amount Reactant Side | Element | Amount on Product Side |
|----------------------|---------|-------------------------|
| 5 | C | 5 |
| 12 | H | 12 |
| 16 | O | 16 |
**Status:**
- The equation is balanced, as indicated by the check mark and the word "balanced" written next to the table.
**Navigation:**
- Arrows indicate options to "Go back," "Click here for a hint," "Click here to check your answer," or "Go to next question."
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