Question
Balance the following equation:
C3H8+O2 = CO+H2O
Select the coefficient that would go in front of O2
C3H8+O2 = CO+H2O
Select the coefficient that would go in front of O2
Answers
GPT-4o mini
To balance the equation for the combustion of propane (C3H8), we start with the unbalanced equation:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO} + \text{H}_2\text{O} \]
1. **Count the number of atoms on each side:**
- Left side: 3 Carbon (C), 8 Hydrogen (H), and an unknown number of Oxygen (O).
- Right side: 1 Carbon (from CO), 2 Hydrogen (from H2O), and 1 Oxygen (from CO) + 1 Oxygen (from H2O) = 2 Oxygen.
2. **Balance the carbon atoms:**
- We have 3 carbons in C3H8, so we need 3 CO on the right:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO} + \text{H}_2\text{O} \]
3. **Balance the hydrogen atoms:**
- We have 8 hydrogens in C3H8, so we need 4 H2O on the right:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO} + 4 \text{H}_2\text{O} \]
4. **Count the number of oxygen atoms needed on the right:**
- From 3 CO, we have 3 O.
- From 4 H2O, we have 4 O.
- Total oxygen on the right: 3 + 4 = 7 O.
5. **Balance the oxygen atoms:**
- Since O2 has 2 oxygen atoms per molecule, to get 7 O total, we need \(\frac{7}{2}\) O2:
\[ \text{C}_3\text{H}_8 + \frac{7}{2} \text{O}_2 \rightarrow 3 \text{CO} + 4 \text{H}_2\text{O} \]
6. **Convert to whole number coefficients:**
- Multiplying the entire equation by 2 to eliminate the fraction gives:
\[ 2 \text{C}_3\text{H}_8 + 7 \text{O}_2 \rightarrow 6 \text{CO} + 8 \text{H}_2\text{O} \]
Thus, the coefficient that would go in front of O2 is **7**.
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO} + \text{H}_2\text{O} \]
1. **Count the number of atoms on each side:**
- Left side: 3 Carbon (C), 8 Hydrogen (H), and an unknown number of Oxygen (O).
- Right side: 1 Carbon (from CO), 2 Hydrogen (from H2O), and 1 Oxygen (from CO) + 1 Oxygen (from H2O) = 2 Oxygen.
2. **Balance the carbon atoms:**
- We have 3 carbons in C3H8, so we need 3 CO on the right:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO} + \text{H}_2\text{O} \]
3. **Balance the hydrogen atoms:**
- We have 8 hydrogens in C3H8, so we need 4 H2O on the right:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO} + 4 \text{H}_2\text{O} \]
4. **Count the number of oxygen atoms needed on the right:**
- From 3 CO, we have 3 O.
- From 4 H2O, we have 4 O.
- Total oxygen on the right: 3 + 4 = 7 O.
5. **Balance the oxygen atoms:**
- Since O2 has 2 oxygen atoms per molecule, to get 7 O total, we need \(\frac{7}{2}\) O2:
\[ \text{C}_3\text{H}_8 + \frac{7}{2} \text{O}_2 \rightarrow 3 \text{CO} + 4 \text{H}_2\text{O} \]
6. **Convert to whole number coefficients:**
- Multiplying the entire equation by 2 to eliminate the fraction gives:
\[ 2 \text{C}_3\text{H}_8 + 7 \text{O}_2 \rightarrow 6 \text{CO} + 8 \text{H}_2\text{O} \]
Thus, the coefficient that would go in front of O2 is **7**.