Certainly! Let's go through the process of balancing the chemical equation for the combustion of ethanol:
### Given Unbalanced Equation:
[tex]\[ - \text{C}_2 \text{H}_5 \text{OH} + \ldots \text{O}_2 \rightarrow - \text{CO}_2 + - \text{H}_2 \text{O} \][/tex]
### Step-by-Step Solution:
1. Write down the unbalanced equation:
[tex]\[ \text{C}_2 \text{H}_5 \text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2 \text{O} \][/tex]
2. Balance the carbon (C) atoms:
- There are 2 carbon atoms in ethanol [tex]\((\text{C}_2 \text{H}_5 \text{OH})\)[/tex].
- Each carbon dioxide molecule ([tex]\(\text{CO}_2\)[/tex]) contains 1 carbon atom.
- Therefore, we need 2 molecules of carbon dioxide to balance the carbon atoms.
[tex]\[ \text{C}_2 \text{H}_5 \text{OH} + \text{O}_2 \rightarrow 2 \text{CO}_2 + \text{H}_2 \text{O} \][/tex]
3. Balance the hydrogen (H) atoms:
- There are 6 hydrogen atoms in ethanol [tex]\((\text{C}_2 \text{H}_5 \text{OH})\)[/tex].
- Each water molecule ([tex]\(\text{H}_2 \text{O}\)[/tex]) contains 2 hydrogen atoms.
- Therefore, we need 3 molecules of water to balance the hydrogen atoms.
[tex]\[ \text{C}_2 \text{H}_5 \text{OH} + \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2 \text{O} \][/tex]
4. Balance the oxygen (O) atoms:
- On the right-hand side, there are:
- From [tex]\( \text{CO}_2 \)[/tex]: [tex]\(2 \times 2 = 4\)[/tex] oxygen atoms.
- From [tex]\( \text{H}_2 \text{O} \)[/tex]: [tex]\(3 \times 1 = 3\)[/tex] oxygen atoms.
- Total oxygen atoms needed on the right-hand side = [tex]\(4 + 3 = 7\)[/tex] oxygen atoms.
- On the left-hand side, there are:
- 1 oxygen atom in the ethanol [tex]\((\text{C}_2 \text{H}_5 \text{OH})\)[/tex].
- So we've already got 1 oxygen atom from ethanol, we need [tex]\(7 - 1 = 6\)[/tex] more oxygen atoms from [tex]\(\text{O}_2\)[/tex].
- Each [tex]\(\text{O}_2\)[/tex] molecule provides 2 oxygen atoms, so we need [tex]\(6 \div 2 = 3\)[/tex] [tex]\(\text{O}_2\)[/tex] molecules.
[tex]\[ \text{C}_2 \text{H}_5 \text{OH} + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2 \text{O} \][/tex]
5. Avoid fractional coefficients:
- To avoid fractional coefficients, multiply all coefficients by 2:
[tex]\[ 2 \text{C}_2 \text{H}_5 \text{OH} + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2 \text{O} \][/tex]
### Resulting Balanced Equation:
[tex]\[ \boxed{2 \text{C}_2 \text{H}_5 \text{OH} + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2 \text{O}} \][/tex]
Now, if you consider the coefficients from the problem statement:
- Coefficient of [tex]\(\text{C}_2 \text{H}_5 \text{OH}\)[/tex] is -2
- Coefficient of [tex]\(\text{O}_2\)[/tex] is 5
- Coefficient of [tex]\(\text{CO}_2\)[/tex] is 4
- Coefficient of [tex]\(\text{H}_2 \text{O}\)[/tex] is 6
So, incorporating these coefficients, we would obtain:
[tex]\[ \boxed{-2 \text{C}_2 \text{H}_5 \text{OH} + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2 \text{O}} \][/tex]
