Write a balanced chemical equation based on the following description: butanoic acid, C3H,COOH(I) burns in air

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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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**Question 4 of 22**

Write a balanced chemical equation based on the following description: butanoic acid, C₃H₇COOH(l) burns in air.

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**Explanation:**

The prompt asks for the chemical reaction when butanoic acid burns in air, which typically involves a combustion reaction. In a combustion reaction, a substance reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). 

Steps to balance the equation:

1. Write the unbalanced equation:
   \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \]

2. Balance carbon atoms:
   \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow 4\text{CO}_2(g) + \text{H}_2\text{O}(l) \]

3. Balance hydrogen atoms:
   \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \]

4. Balance oxygen atoms:
   - Total oxygen atoms needed on the product side: \(4 \times 2 (\text{from CO}_2) + 4 \times 1 (\text{from H}_2\text{O}) = 12\)
   - Oxygen in butanoic acid: 2 (from C₃H₇COOH)
   - Oxygen needed from O₂: \( \frac{10}{2} = 5 \)

5. Final balanced equation:
   \[ \text{C}_3\text{H}_7\text{COOH(l)} + \frac{9}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \]

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**Note:**
To avoid using fractional coefficients, multiply the entire equation by 2:

\[ 2\text{C}_3\text
Transcribed Image Text:**Question 4 of 22** Write a balanced chemical equation based on the following description: butanoic acid, C₃H₇COOH(l) burns in air. --- **Explanation:** The prompt asks for the chemical reaction when butanoic acid burns in air, which typically involves a combustion reaction. In a combustion reaction, a substance reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). Steps to balance the equation: 1. Write the unbalanced equation: \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \] 2. Balance carbon atoms: \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow 4\text{CO}_2(g) + \text{H}_2\text{O}(l) \] 3. Balance hydrogen atoms: \[ \text{C}_3\text{H}_7\text{COOH(l)} + \text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \] 4. Balance oxygen atoms: - Total oxygen atoms needed on the product side: \(4 \times 2 (\text{from CO}_2) + 4 \times 1 (\text{from H}_2\text{O}) = 12\) - Oxygen in butanoic acid: 2 (from C₃H₇COOH) - Oxygen needed from O₂: \( \frac{10}{2} = 5 \) 5. Final balanced equation: \[ \text{C}_3\text{H}_7\text{COOH(l)} + \frac{9}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \] --- **Note:** To avoid using fractional coefficients, multiply the entire equation by 2: \[ 2\text{C}_3\text
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