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.
