Balancing Combustion Reactions

We'll look at three different combustion reactions to see some of the common patterns that occur.

Pattern 1 - Easy

Here's the first example:

The first thing to know is that you should balance the oxygen last.

This is not because oxygen is somehow special, it is because it is alone on the left. In other words, oxygen is the only element in this reaction that can be balanced (with a coefficient) without changing at least two different element amounts. (If you put a coefficient in front of CH₄, you alter both C and H, etc.)

In this case there is one C on the left and one on the right...so far so good. There are 4 H atoms on the left, but two on the right. That can be fixed easily:




Now both C and H are balanced. There are 2 O atoms on the left, and 4 on the right (2 in the CO2 and 2 in the 2 water molecules). Again, an easy fix.

Pattern 2 - When in trouble...

Lots of combustion reactions work this neatly. Some however are more difficult. Here's an example that shows a common issue with combustion reactions:



Two carbon atoms on the left and one on the right – that can be easily fixed.



There are 6 H atoms on the left and 2 on the right. This, also, can be easily fixed.



Now, however, we have a problem. There are 7 oxygen atoms on the right (4 from the two CO₂ molecules and 3 more from the water), but the oxygen comes in pairs (O₂ on the left).

There are LOTS of ways to think your way out of this, but the easiest is the simple advice “When in trouble...double.” In other words, when you find an odd/even issue, try doubling everything (at least everything that you've already dealt with).



Now, if we re-count the oxygen atoms we have 14 on the right (8 from the CO₂'s and 6 from the waters). This is easy to work out.

Pattern 3 - Splitting the oxygens...

A last example of combustion balancing:



We can start with the C's and quickly get to this point:



The first stumbling block is to mis-count the H atoms on the left side of the reaction. There are 6, not 5. Even though they are written separately, they all count together. That leads us to:



If we count oxygen on the right side we get 7 (just like last time) and so you might be tempted to think that this reaction needs to be doubled. But...oxygen is in BOTH reactants, not just the O₂. So, ONE of the O atoms comes from the C₂H₅OH, the rest of them (the other 6 needed to make 7) must come from the O₂.

That means that we need 3 oxygen molecules: