# Question #cf542

Mar 15, 2016

The balanced equation is $\text{2KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O}$

#### Explanation:

The procedure I usually use to balance equations like this is:

2. Balance all atoms other than $\text{H}$ and $\text{O}$.
3. Balance $\text{O}$.
4. Balance $\text{H}$.
5. If necessary, multiply by an integer to remove fractions.
6. Check that all atoms are balanced.

$\text{KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O}$

Put a 1 in front of ${\text{K"_2"CO}}_{3}$. This number is fixed and does not change until (if necessary) Step 5.

$\text{KOH"color(white)(l) + "CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O}$

Step 2. Balance $\text{K}$.

We have fixed $\text{2 K}$ on the right, so we need $\text{2 K}$ on the left.

Put a 2 before $\text{KOH}$.

$\textcolor{b l u e}{2} \text{KOH" + "CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O}$

Step 3. Balance $\text{C}$.

We have fixed $\text{1 C}$ on the right, so we need $\text{1 C}$ on the left.

Put a 1 before ${\text{CO}}_{2}$.

$\textcolor{b l u e}{2} \text{KOH" + color(orange)(1)"CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O}$

-+++++++++++++3
·······

Step 4. Balance $\text{O}$.

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

Put a 1 in front of $\text{H"_2"O}$.

$\textcolor{b l u e}{2} \text{KOH" + color(orange)(1)"CO"_2 → color(red)(1)"K"_2"CO"_3 + color(green)(1)"H"_2"O}$

All formulas have a coefficient, so the equation should be balanced.

Step 5.

(Not needed.)

Step 6. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l) bb"On the right}}$
$\textcolor{w h i t e}{m l l} \text{2 K"color(white)(mmmmml) "2 K}$
$\textcolor{w h i t e}{m l l} \text{4 O"color(white)(mmmmml) "4 O}$
$\textcolor{w h i t e}{m l l} \text{2 H"color(white)(mmmmml) "2 H}$
$\textcolor{w h i t e}{m l l} \text{1 C"color(white)(mmmmml) "1 C}$

The balanced equation is

$\textcolor{red}{\text{2KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O}}$