# How do you balance C_2H_6 + O_2 -> CH_3COOH + H_2O?

Jun 9, 2016

${C}_{2} {H}_{6} + \frac{3}{2} {O}_{2} \rightarrow C {H}_{3} C O O H + {H}_{2} O$ or $2 {C}_{2} {H}_{6} + 3 {O}_{2} \rightarrow 2 C {H}_{3} C O O H + 2 {H}_{2} O$

#### Explanation:

First you want to do an atom inventory. Determine the number of atoms of each element that are on both side of the reaction arrow:

Reactants side:
C atoms = 2
H atoms = 6
O atoms = 2

Products side:
C atoms = 2
H atoms = 6
O atoms = 3

As we can see the only element that isn't balanced is oxygen. A simple approach to this would be to focus on the ${O}_{2}$ on the reactants side. You want to find a coefficient that would make the number of oxygen atoms on both side to be the same.

Place a coefficient of 3/2 in front of ${O}_{2}$ so the total number of atoms will equal three since you always multiply the coefficient by the subscript that's attached to that atom.

Then you'll end up with this ${C}_{2} {H}_{6} + \frac{3}{2} {O}_{2} \rightarrow C {H}_{3} C O O H + {H}_{2} O$

Since some people don't like the idea of fractional coefficients, you could multiply the entire chemical reaction by 2 to obtain whole number coefficients like so:

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

I hope this makes sense!