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

1 Answer
Jun 9, 2016

Answer:

#C_2H_6 + 3/2O_2 rarr CH_3COOH + H_2O# or #2C_2H_6 + 3O_2 rarr 2CH_3COOH + 2H_2O#

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_2H_6 + 3/2O_2 rarr CH_3COOH + H_2O#

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:

#2C_2H_6 + 3O_2 rarr 2CH_3COOH + 2H_2O#

I hope this makes sense!

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