# Can you write and balance the equation for the complete combustion of ethane, C_2H_6?

Nov 12, 2015

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

#### Explanation:

In a combustion reaction with a hydrocarbon in the reactant side you will always have ${O}_{2}$ as another reactant. As you will always have $C {O}_{2}$ and ${H}_{2} O$ as the products.

Knowing that much you can set up your reaction equation..

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

Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.

Start with the $C$ atoms first and move to the $H$ atoms next. It's easier to leave the ${O}_{2}$ to the last, it has a way to alter the equation.

Initially, you would arrive at this, before the ${O}_{2}$ has been balanced:

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

But, as you can see, you have an odd amount of ${O}_{2}$ on the product side. In this case, you have to find the common factor of the amount of $O$ on the product side and 2, Because of the ${O}_{2}$ diatom. Therefore, 14 would be the lowest common factor of 2 and 7.

Jan 10, 2016

${C}_{2} {H}_{6} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(g\right)$