# How would you balance the following equation: CH3CH2CH3(g)+O2(g) --> CO2(g) + H2O(g)?

Dec 4, 2015

We can rewrite this as:

$\text{C"_3"H"_8(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O} \left(g\right)$

I would initially start with the number of carbons because the number of carbons in ${\text{CO}}_{2}$ is even, and the number of oxygens in ${\text{O}}_{2}$ is also even.

$\text{C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + "H"_2"O} \left(g\right)$

Then, I would realize that there are $8$ hydrogens on the left, and I need $4 \times 2$ on the right.

$\text{C"_3"H"_8(g) + "O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O} \left(g\right)$

Lastly, I would balance the oxygen, because it is easy to only affect the number of oxygens. It only complicates things more to assign a number to ${\text{O}}_{2}$ first because you'll probably end up changing it again later.

$\textcolor{b l u e}{\text{C"_3"H"_8(g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O} \left(g\right)}$

Last confirmation:

$\text{C}$: $1 \times 3 = 3 \times 1$
$\text{H}$: $1 \times 8 = 4 \times 2$
$\text{O}$: $5 \times 2 = 3 \times 2 + 4 \times 1$

Since this cannot be reduced any further, this is good to go.