What is the balanced equation for the combustion of propan-1-ol?

2 CH3CH2CH2OH + 9 O2 >> 6 CO2 + 8H2O

Propanol is C3H8O. When you burn it, each C turns into CO2, and every 2 H's turn into an H2O molecule. The rest of the balancing involves the O atoms. Without balancing the O's, you get:

C3H8O + xxx O2 --> 3 CO2 + 4 H2O. Ignoring the contribution from O2, there is one O on the left, and there are 10 on the right. That means 9 O atoms must be added to the left. Oxygen exists as O2, so 9 O atoms are equivalent to 4.5 O2's.

The equation:

C3H8O + 4.5 O2 --> 3 CO2 + 4 H2O

is balanced. It just doesn't have all of the coefficients as whole numbers, so it will be fixed by multiplying everything by two:

2x(C3H8O + 4.5 O2 --> 3 CO2 + 4 H2O)
= 2 C3H8O + 9 O2 --> 6 CO2 + 8 H2O.

2 C3H7OH + 9 O2 >> 6 CO2 + 8H2O + heat

C3H7OH = CH3CH2CH2OH

Propan-1-ol