Balanced equation for the reaction between propane and insufficient oxygen

There are 3 carbon atoms on the left, so we need 3 carbon atoms on the right.

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

There are 8 hydrogen atoms on the left, so we need 8 hydrogen atoms on the right. Since water has 2 hydrogen atoms per molecule, we need $8/2 = 4$ molecules of water.

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

Count the oxygen atoms on the right side: $3 \times 1 (\text{from CO}) + 4 \times 1 (\text{from H}_2\text{O}) = 3 + 4 = 7$ oxygen atoms.
Since oxygen on the left is in $\text{O}_2$ molecules, we need $7/2$ molecules of $\text{O}_2$.

$$\text{C}_3\text{H}_8 + \frac{7}{2}\text{O}_2 \rightarrow 3\text{CO} + 4\text{H}_2\text{O}$$

$$2\text{C}_3\text{H}_8 + 7\text{O}_2 \rightarrow 6\text{CO} + 8\text{H}_2\text{O}$$

The balanced equation for the reaction between propane and insufficient oxygen (producing carbon monoxide) is:

$$\boxed{2\text{C}_3\text{H}_8 + 7\text{O}_2 \rightarrow 6\text{CO} + 8\text{H}_2\text{O}}$$