# How do you balance C_3H_6 + O_2 -> CO + H_2O?

Jan 29, 2016

${C}_{3} {H}_{6} + 3 {O}_{2} \to 3 C O + 3 {H}_{2} O$

#### Explanation:

To balance, we add molecules such as to ensure that the number of atoms of each element is the same on each side of the equation.

This equation represents the incomplete combustion of propane as insufficient oxygen must have been present since carbon monoxide was form instead of carbon dioxide which would form if there was sufficient oxygen to ensure complete combustion.

Jan 29, 2016

${C}_{3} {H}_{6} + 3 {O}_{2} \rightarrow 3 C O + 3 {H}_{2} O$

#### Explanation:

We have three carbons on the reagents' side and only one on the products' side, so we need to balance that by making that equal.

${C}_{3} {H}_{6} + {O}_{2} \rightarrow 3 C O + {H}_{2} O$

We have six hydrogens on the reagents' side and only 2 on the products' side so we need to balance that by making them equal.

${C}_{3} {H}_{6} + {O}_{2} \rightarrow 3 C O + 3 {H}_{2} O$

We have 2 oxygens on the reagents' side and 6 on the products' side, so we need to balance that by making them equal, now, since the reagent's side has only one substance with oxygen it'll be easier to make that equal, so,

${C}_{3} {H}_{6} + 3 {O}_{2} \rightarrow 3 C O + 3 {H}_{2} O$

Do note that if you had equalized the first time around by

$\frac{1}{3} {C}_{3} {H}_{6} + {O}_{2} \rightarrow C O + {H}_{2} O$

It'd be already balanced, but it's considered good practice to balance equations by having integer coefficients, which we could have solved by multiplying everything by 3

${C}_{3} {H}_{6} + 3 {O}_{2} \rightarrow 3 C O + 3 {H}_{2} O$