A combustion reaction will only produce carbon dioxide, CO_2, and water, H_2O. An unbalanced chemical equation for the combustion of a general hydrocarbon, C_xH_y, wil be
C_xH_y + O_2 -> CO_2 + H_2O
Regardless of what values you have for x and y, the balanced chemical equation will always have these stoichiometric coefficients
C_xH_y + (x+y/4)O_2 -> xCO_2 + y/2H_2O
Take, for example, the combustion of methane, CH_4. For methane, you have x=1 and y=4. PLug these values into the equation above and you'll get
C_1H_4 + (1 + 4/4)O_2 ->CO_2 + 4/2H_2O
which ends up being
CH_4 + 2O_2 -> CO_2 + 2H_2O -> balanced
Another example could be the combustion of ethane, C_2H_6, for which x=2 and y=6.
C_2H_6 + (2 + 6/4)O_2 -> 2CO_2 + 6/2H_2O
which ends up being
C_2H_6 + 7/2O_2 -> 2CO_2 + 3H_2O
If you're not allowed to use fractional coefficients, multiply all the coefficients by 2 to get
2C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O -> balanced
That's how you balance the combustion reaction of any hydrocarbon.
