# Question #dc820

May 9, 2015

A combustion reaction will only produce carbon dioxide, $C {O}_{2}$, and water, ${H}_{2} O$. An unbalanced chemical equation for the combustion of a general hydrocarbon, ${C}_{x} {H}_{y}$, wil be

${C}_{x} {H}_{y} + {O}_{2} \to C {O}_{2} + {H}_{2} O$

Regardless of what values you have for $x$ and $y$, the balanced chemical equation will always have these stoichiometric coefficients

${C}_{x} {H}_{y} + \left(x + \frac{y}{4}\right) {O}_{2} \to x C {O}_{2} + \frac{y}{2} {H}_{2} O$

Take, for example, the combustion of methane, $C {H}_{4}$. For methane, you have $x = 1$ and $y = 4$. PLug these values into the equation above and you'll get

${C}_{1} {H}_{4} + \left(1 + \frac{4}{4}\right) {O}_{2} \to C {O}_{2} + \frac{4}{2} {H}_{2} O$

which ends up being

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$ $\to$ balanced

Another example could be the combustion of ethane, ${C}_{2} {H}_{6}$, for which $x = 2$ and $y = 6$.

${C}_{2} {H}_{6} + \left(2 + \frac{6}{4}\right) {O}_{2} \to 2 C {O}_{2} + \frac{6}{2} {H}_{2} O$

which ends up being

${C}_{2} {H}_{6} + \frac{7}{2} {O}_{2} \to 2 C {O}_{2} + 3 {H}_{2} O$

If you're not allowed to use fractional coefficients, multiply all the coefficients by 2 to get

$2 {C}_{2} {H}_{6} + 7 {O}_{2} \to 4 C {O}_{2} + 6 {H}_{2} O$ $\to$ balanced

That's how you balance the combustion reaction of any hydrocarbon.