Question

How to balance the equation for the combustion of propane

Answer 1

To balance the equation for the combustion of propane (C₃H₈), we start with the unbalanced equation:

\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]

1. **Count the number of atoms of each element** in the reactants and products.

- Reactants: C = 3, H = 8, O = 2 (in O₂)
- Products: C = 1 (in CO₂), H = 2 (in H₂O)

2. **Balance the carbon atoms** first. There are 3 carbon atoms in propane, so we need 3 CO₂:

\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O}
\]

3. **Balance the hydrogen atoms** next. There are 8 hydrogen atoms in propane, which means we need 4 H₂O to balance the hydrogen:

\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]

4. **Count the oxygen atoms** now. We have:
- From 3 CO₂: \(3 \times 2 = 6\) O
- From 4 H₂O: \(4 \times 1 = 4\) O
- Total oxygen atoms needed: \(6 + 4 = 10\) O

5. On the left side, we have O₂. Since each O₂ molecule contains 2 oxygen atoms, we need 5 O₂ to provide 10 oxygen atoms:

\[
5 \text{O}_2
\]

6. Now we can write the fully balanced equation:

\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]

This equation is now balanced with:

- **Carbons:** 3 on both sides
- **Hydrogens:** 8 on both sides
- **Oxygens:** 10 on both sides

Therefore, the balanced equation for the combustion of propane is:

\[
\text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O}
\]