# How do you balance CH3OH + O2 -- CO2 + H2O?

Oct 30, 2016

The balanced equation is $2 \text{CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O}$

#### Explanation:

You follow a systematic procedure to balance the equation.

$\text{CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O}$

A method that often works is first to balance everything other than $\text{O}$ and $\text{H}$, then balance $\text{O}$, and finally balance $\text{H}$.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like $\text{CH"_3"OH}$. We put a 1 in front of it to remind ourselves that the number is now fixed.

$\textcolor{red}{1} \text{CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O}$

Balance $\text{C}$:

We have $\text{1 C}$ on the left, so we need $\text{1 C}$ on the right. We put a 1 in front of the ${\text{CO}}_{2}$.

$\textcolor{red}{1} \text{CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O}$

Balance $\text{H}$:

We can't balance $\text{O}$ because we have two oxygen-containing molecules without coefficients. ∴ Let's balance $\text{H}$ instead.

We have $\text{4 H}$ on the left, so we need $\text{4 H}$ on the right. There are already $\text{2 H}$ atoms on the right. We must put a 2 in front of the $\text{H"_2"O}$.

$\textcolor{red}{1} \text{CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O}$

Balance $\text{O}$:

We have fixed $\text{4 O}$ on the right and $\text{1 O}$ on the left. We need $\text{3 O}$ on the left.

Uh, oh! Fractions!

We start over, this time doubling all the coefficients.

$\textcolor{red}{2} \text{CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O}$

Now we can balance $\text{O}$ by putting a 3 in front of ${\text{O}}_{2}$

$\textcolor{red}{2} \text{CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O}$

Every formula now has a coefficient. We should have a balanced equation.

Let's check.

$\text{Atom" color(white)(m)"lhs"color(white)(m)"rhs}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m l} 2 \textcolor{w h i t e}{m m} 2$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m l} 8 \textcolor{w h i t e}{m m} 8$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m l} 8 \textcolor{w h i t e}{m m} 8$

All atoms balance. The balanced equation is

$2 \text{CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O}$

Oct 31, 2016

$2 C {H}_{3} O H$ $+ 3 {O}_{2}$ $\rightarrow 2 C {O}_{2} + 4 {H}_{2} O$ .

#### Explanation:

• First try to balance atoms other than O and H. So first balance the C atoms.
$2 C {H}_{3} O H$ $+ {O}_{2}$ $\rightarrow 2 C {O}_{2} + {H}_{2} O$ .
• Then balance the H atoms, as no. of H atoms is less than no. of O atoms.
$2 C {H}_{3} O H$ $+ {O}_{2}$ $\rightarrow 2 C {O}_{2} + 4 {H}_{2} O$ .
• Now, balance the remaining O atoms.
$2 C {H}_{3} O H$ $+ 3 {O}_{2}$ $\rightarrow 2 C {O}_{2} + 4 {H}_{2} O$ .

Now, check that the no. of all individual atoms in L.H.S and R.H.S are equal.