To balance the chemical equation for the combustion of propane ([tex]\( \text{CH}_3\text{CH}_2\text{CH}_3 \)[/tex]), we need to ensure that the number of each type of atom on the reactants side equals the number of each type on the products side. The unbalanced equation is:
[tex]\[
\text{CH}_3\text{CH}_2\text{CH}_3(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g)
\][/tex]
We need to balance the carbon, hydrogen, and oxygen atoms. Let's go step-by-step:
1. Balance Carbon (C) atoms:
[tex]\(\text{CH}_3\text{CH}_2\text{CH}_3\)[/tex] contains 3 carbon atoms. Therefore, we need 3 [tex]\(\text{CO}_2\)[/tex] molecules to balance the carbon on both sides.
[tex]\[
\text{CH}_3\text{CH}_2\text{CH}_3(g) + \text{O}_2(g) \rightarrow 3\text{CO}_2(g) + \text{H}_2\text{O}(g)
\][/tex]
2. Balance Hydrogen (H) atoms:
[tex]\(\text{CH}_3\text{CH}_2\text{CH}_3\)[/tex] contains 8 hydrogen atoms. Therefore, we need 4 [tex]\(\text{H}_2\text{O}\)[/tex] molecules to balance the hydrogen on both sides (since each [tex]\(\text{H}_2\text{O}\)[/tex] molecule contains 2 hydrogen atoms).
[tex]\[
\text{CH}_3\text{CH}_2\text{CH}_3(g) + \text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
\][/tex]
3. Balance Oxygen (O) atoms:
On the products side, we have:
- 3 [tex]\(\text{CO}_2\)[/tex] molecules contributing [tex]\(3 \times 2 = 6\)[/tex] oxygen atoms.
- 4 [tex]\(\text{H}_2\text{O}\)[/tex] molecules contributing [tex]\(4 \times 1 = 4\)[/tex] oxygen atoms.
This gives a total of [tex]\(6 + 4 = 10\)[/tex] oxygen atoms on the products side. Since [tex]\(\text{O}_2\)[/tex] is a diatomic molecule, each [tex]\(\text{O}_2\)[/tex] molecule provides 2 oxygen atoms. Therefore, we need [tex]\( \frac{10}{2} = 5 \)[/tex] [tex]\(\text{O}_2\)[/tex] molecules to balance the oxygen atoms.
[tex]\[
\text{CH}_3\text{CH}_2\text{CH}_3(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
\][/tex]
So, the balanced chemical equation is:
[tex]\[
\text{CH}_3\text{CH}_2\text{CH}_3(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
\][/tex]
Double-checking:
- Carbon: [tex]\(1 \text{ molecule of } \text{CH}_3\text{CH}_2\text{CH}_3 \)[/tex] = 3 atoms of C [tex]\(\rightarrow\)[/tex] [tex]\(3 \text{ molecules of } \text{CO}_2 \)[/tex] = 3 atoms of C.
- Hydrogen: [tex]\(1 \text{ molecule of } \text{CH}_3\text{CH}_2\text{CH}_3 \)[/tex] = 8 atoms of H [tex]\(\rightarrow\)[/tex] [tex]\(4 \text{ molecules of } \text{H}_2\text{O} \)[/tex] = 8 atoms of H.
- Oxygen: [tex]\(5 \text{ molecules of } \text{O}_2 \)[/tex]= 10 atoms of O [tex]\(\rightarrow\)[/tex] [tex]\(3 \text{ molecules of } \text{CO}_2 = 6\)[/tex] atoms of O and [tex]\(4 \text{ molecules of } \text{H}_2\text{O} = 4\)[/tex] atoms of O (6 + 4 = 10 atoms of O).
The balancing is correct.
