To balance the combustion reaction of propane (C₃H₈), let's carefully follow these steps to ensure the number of each type of atom on the reactants side is equal to the number on the products side.
1. Write the unbalanced equation:
[tex]\( \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)[/tex]
2. Balance the carbon atoms first:
- There are 3 carbon (C) atoms in propane (C₃H₈).
- To balance the carbons, we need 3 molecules of carbon dioxide (CO₂).
[tex]\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O}
\][/tex]
3. Balance the hydrogen atoms next:
- There are 8 hydrogen (H) atoms in propane (C₃H₈).
- To balance the hydrogens, we need 4 molecules of water (H₂O), because each molecule of water has 2 hydrogen atoms.
[tex]\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\][/tex]
4. Balance the oxygen atoms last:
- On the product side, we have:
- 3 molecules of CO₂, each contributing 2 oxygen atoms: [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms.
- 4 molecules of H₂O, each contributing 1 oxygen atom: [tex]\( 4 \times 1 = 4 \)[/tex] oxygen atoms.
- So, the total oxygen atoms needed on the reactants side is [tex]\( 6 + 4 = 10 \)[/tex] oxygen atoms.
- Since each O₂ molecule contains 2 oxygen atoms, we need [tex]\( \frac{10}{2} = 5 \)[/tex] O₂ molecules.
[tex]\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\][/tex]
Therefore, the balanced chemical equation is:
[tex]\[ \boxed{\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}} \][/tex]
