CH3CH2OH + O2 → CO2 + H2O
The reaction above is combustion of a compound with C, H, and O.
The following method will balance any combustion equation of CxHyOz.
1st balance the C’s
2 C’s on left, 2 C’s on right. Make 2 C’s on right, by making 2 CO2’s on right.
CH3CH2OH + O2 → 2 CO2 + H2O
2nd balance the H’s
6 H’s on left, 2 H’s on right. Make 6 H’s on right, by making 3 H2O’s on right.
CH3CH2OH + O2 → 2 CO2 + 3 H2O
3rd balance the O’s
You do not want to change the coefficient of CH3CH2OH. So, balance the O’s on the right by changing the coefficient of O2.
4 O’s + 3 O’s = 7 O’s on right. Make 3 O2’s on left.
CH3CH2OH + 3 O2 → 2 CO2 + 3 H2O
Check:
2 C’s = 2 C’s
3 + 2 + 1 H’s = 3 * 2 H’s
1 + (3 * 2) O’s = (2 * 2) + 3 O’s
If the coefficient of O2 is a fraction, you multiply both sides of the equation by the denominator of the fraction.
Example:
C4H10 + O2 → CO2 + H2O
The coefficient of CO2 is equal to the number of C’s in the reactant = 4
C4H10 + O2 → 4 CO2 + H2O
The coefficient of H2O is equal to ½ of the number of H’s in the reactant = 5
C4H10 + O2 → 4 CO2 + 5 H2O
Right side has 8 + 5 = 13 O’s. Make 6.5 O2.
C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
4th double the entire equation so all coefficients are whole numbers
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
