# How to balance chemical equations using algebraic method?

Jan 18, 2017

Here's how you do it.

#### Explanation:

Assume you must balance the equation

${\text{CaCO"_3 + "H"_3"PO"_4 → "Ca"_3("PO"_4)_2 + "H"_2"CO}}_{3}$

You first write the equation using letter variables for the coefficients:

$a {\text{CaCO"_3 + b"H"_3"PO"_4 → c"Ca"_3("PO"_4)_2 + d"H"_2"CO}}_{3}$

Then you set up a series of simultaneous equations, one for each element.

$\text{Ca} : \textcolor{w h i t e}{m l} a \textcolor{w h i t e}{m m l} = 3 c$
$\text{C} : \textcolor{w h i t e}{m l l} a \textcolor{w h i t e}{m m l l} = \textcolor{w h i t e}{m m m} d$
$\text{O} : \textcolor{w h i t e}{m} 3 a + 4 b = 8 c + 3 d$
$\text{H} : \textcolor{w h i t e}{m m m l l} 3 b = \textcolor{w h i t e}{m m l l} 2 d$
$\text{P} : \textcolor{w h i t e}{m m m m l} b = 2 c$

Now you solve the five simultaneous equations.

It looks almost impossible, but we know that the coefficients must be integers.

Let's set $c = 1$.

Then $\textcolor{w h i t e}{l} a = 3$ and

$\textcolor{w h i t e}{m m l l} d = a = 3$

$\textcolor{w h i t e}{m m l l} b = 2 c = 2$

So $a = 3$; $b = 2$; $c = 1$; $d = 3$

The balanced equation is

${\text{3CaCO"_3 + "2H"_3"PO"_4 → "Ca"_3("PO"_4)_2 + "3H"_2"CO}}_{3}$

Here's a useful equation on balancing equations by the algebraic method.