Question
Balance the combustion equation for propane () burning in oxygen:
Solution — Step by Step
Left side: 3 carbon, 8 hydrogen, 2 oxygen (in ).
Right side: 1 carbon (in ), 2 hydrogen + 1 oxygen (in ), 2 oxygen (in ).
We have an imbalance everywhere — let’s fix carbon first, then hydrogen, then oxygen last.
We have 3 carbons on the left. Each has 1 carbon, so we need 3 on the right:
Carbon is now balanced: 3 = 3. ✓
We have 8 hydrogens on the left. Each has 2 hydrogens, so we need 4 :
Hydrogen is now balanced: 8 = 8. ✓
Right side oxygen: (from ) + (from ) = 10 oxygen atoms.
Each provides 2 oxygen atoms, so we need molecules of :
| Element | Left | Right |
|---|---|---|
| C | 3 | 3 ✓ |
| H | 8 | 8 ✓ |
| O | 10 | 10 ✓ |
Balanced equation:
Why This Works
We balanced in order: C → H → O. This ordering is strategic. Carbon and hydrogen appear only in one compound each on the right ( and ), so we can fix them directly. Oxygen, on the other hand, appears in multiple compounds — , , and . So we adjust oxygen last, using the free coefficient to make up whatever shortfall remains.
The coefficients (1, 5, 3, 4) represent the mole ratio. In a real combustion: 1 mole of propane needs 5 moles of oxygen to burn completely, producing 3 moles of and 4 moles of water vapour.
Alternative Method — Fraction Method
If oxygen gives a non-integer coefficient, use fractions first, then multiply through:
Start with the right side: needs . Here 5 is already a whole number.
For cases where it isn’t — say you get — write it as first, then multiply the entire equation by 2.
Common Mistake
Balancing oxygen first. Students often see on the left and try to balance oxygen immediately. This forces repeated re-balancing as you fix carbon and hydrogen later. Always save oxygen for last — it’s the most flexible element in combustion reactions because appears only on the left as a pure element.