Combustion Reactions

With plenty of available oxygen, a hydrocarbon burns completely, and every carbon and hydrogen atom ends up fully oxidized: carbon becomes carbon dioxide, hydrogen becomes water. These two products are the most oxygen-rich, most stable forms carbon and hydrogen can reach, so forming them releases the maximum possible energy from the fuel. The water is usually written as a gas, H₂O(g), because the flame is far hotter than 100 °C.

1. Carbon first: every C atom in the fuel becomes one ${\text{CO}}_2$, so the coefficient of ${\text{CO}}_2$ equals the number of C atoms.
2. Hydrogen next: every 2 H atoms become one ${\text{H}}_2\text{O}$, so the coefficient of ${\text{H}}_2\text{O}$ is half the number of H atoms.
3. Oxygen last: add up all O atoms on the product side and divide by 2 to get the coefficient of ${\text{O}}_2$. Because ${\text{O}}_2$ is a free element, changing it disturbs nothing else.
4. If the ${\text{O}}_2$ coefficient is a fraction such as $\frac{13}{2}$, double every coefficient to clear it.
5. If the fuel itself contains oxygen (such as ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$), subtract the fuel's O atoms before dividing by 2.

When oxygen is limited, there isn't enough to fully oxidize every carbon atom to CO₂. Instead, some carbon forms carbon monoxide (CO) — a colourless, odourless, and highly toxic gas — and in severe oxygen shortage, some carbon may not oxidize at all and is released as fine soot (solid carbon particles).

A visible clue to incomplete combustion is a yellow or orange, sooty flame (from glowing unburned carbon particles), compared to the clean blue flame of complete combustion where the fuel-to-oxygen ratio is properly controlled. In the extreme case, carbon is left unoxidized entirely: ${\text{CH}}_4+{\text{O}}_2\to \text{C}+2{\text{H}}_2\text{O}$. Notice that the less oxygen available, the fewer O atoms end up attached to each carbon: ${\text{CO}}_2$ (two), then CO (one), then C (none). Hydrogen always still becomes water.