Let's balance the combustion reaction between propane ([tex]\(\text{C}_3\text{H}_8\)[/tex]) and oxygen ([tex]\(\text{O}_2\)[/tex]). The unbalanced reaction is given as:
[tex]\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\][/tex]
We need to balance the atoms of carbon (C), hydrogen (H), and oxygen (O) on both sides of the equation.
### Step-by-Step Solution:
1. Balance the Carbon atoms:
Propane ([tex]\(\text{C}_3\text{H}_8\)[/tex]) has 3 carbon atoms. Therefore, we will need 3 molecules of carbon dioxide ([tex]\(\text{CO}_2\)[/tex]) on the right side to balance the carbon atoms.
[tex]\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}
\][/tex]
2. Balance the Hydrogen atoms:
Propane has 8 hydrogen atoms, and water ([tex]\(\text{H}_2\text{O}\)[/tex]) has 2 hydrogen atoms per molecule. Therefore, we need 4 water molecules to have 8 hydrogen atoms on the right side.
[tex]\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\][/tex]
3. Balance the Oxygen atoms:
On the right side, we now have:
- 3 molecules of [tex]\(\text{CO}_2\)[/tex], each with 2 oxygen atoms, giving us [tex]\(3 \times 2 = 6\)[/tex] oxygen atoms.
- 4 molecules of [tex]\(\text{H}_2\text{O}\)[/tex], each with 1 oxygen atom, giving us [tex]\(4 \times 1 = 4\)[/tex] oxygen atoms.
This totals [tex]\(6 + 4 = 10\)[/tex] oxygen atoms on the right side.
To get 10 oxygen atoms on the left side, we need 5 molecules of [tex]\(\text{O}_2\)[/tex] since each molecule of [tex]\(\text{O}_2\)[/tex] has 2 oxygen atoms.
[tex]\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\][/tex]
### Final Balanced Equation
The balanced combustion reaction between propane and oxygen is:
[tex]\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\][/tex]
This equation ensures that there are equal numbers of each type of atom on both sides of the reaction, in accordance with the law of conservation of mass.
