# Question #5f14a

Feb 14, 2017

Balance mass, and balance charge.........

#### Explanation:

$\text{H"_3"CCH"_2"OH"+"3O"_2 rarr "2CO"_2 + "3H"_2"O}$

Feb 14, 2017

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

#### Explanation:

The unbalanced chemical equation is

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

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 often is to start with what looks like the most complicated formula.

The most complicated formula looks like $\text{CH"_3"CH"_2"OH}$. We put a 1 in front of it to remind ourselves that the number is now fixed.

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

Balance $\text{C}$:

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

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

Balance $\text{O}$:

We can't balance $\text{O}$ because there are two compounds containing $\text{O}$ that have no coefficients.

Balance $\text{H}$:

We have fixed $\text{6 H}$ on the left. We need $\text{6 H}$ on the right. Put a 3 in front of $\text{H"_2"O}$.

$\textcolor{red}{1} \text{CH"_3"CH"_2"OH" + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(3)"H"_2"O}$

Balance $\text{O}$

We have fixed $\text{7 O}$ atoms on the right and one on the left. We need six more $\text{O}$ atoms on the left. Put a 3 in front of ${\text{O}}_{2}$.

$\textcolor{red}{1} \text{CH"_3"CH"_2"OH" + color(purple)(3)"O"_2 → color(orange)(2)"CO"_2 + color(blue)(3)"H"_2"O}$

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

Let's check.

$\boldsymbol{\text{Atom" color(white)(m)"lhs"color(white)(m)"rhs}}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m l} 2$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m} 6 \textcolor{w h i t e}{m m l} 6$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m} 7 \textcolor{w h i t e}{m m l} 7$

All atoms balance. The balanced equation is

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