Questions regarding balancing reaction equations are frequent. There is a high chance that you probably came across some whilst doing past papers. 

Such question can also be time consuming. However, there is a simpler and more organised way to solve such questions. We can solve such questions using algebraic equations. 

We first need to form algebraic equations for each of the elements present in the equation. 
We need to compare the number of atoms present on the LHS and RHS for each element. 

Let us begin with a simple question. 

Example:
H₂ + aO₂  →  H₂O
Find the value of a.

Hydrogen
Number of atoms on LHS = 2
Number of atoms on RHS = 2
So Hydrogen is already balanced. 

Oxygen
Number of atoms on LHS = a x 2 = 2a
Number of atoms RHS = 1
In order for oxygen atoms to be balanced, we need to satisfy the equation:
2a = 1
Hence, a = ½

The balanced equation would look like:
H₂ + ½O₂  →  H₂O


Complicated Question

Try solving this question on your own first using algebraic equations. 

aCa₃(PO₄)₂ + bSiO₂ + cC  →  dCaSiO₃ + eP₄ + fCO

Calcium
Number of atoms on LHS = 3a
Number of atoms on RHS = d
3a = d —– Eq 1

Phosphorus
Number of atoms on LHS = 2a
Number of atoms on RHS = 4e
2a = 4e
a = 2e —– Eq 2

Oxygen (present in PO₄ nad SiO₂)
Number of atoms on LHS = 8a + 2b
Number of atoms on RHS = 3d + f
8a + 2b = 3d + f —– Eq 3

Silicon
Number of atoms on LHS = b
Number of atoms on RHS = d
b = d —– Eq 4

Carbon
Number of atoms on LHS = c
Number of atoms on RHS = f
c = f —– Eq 5

Let us express everything in terms of ‘a’.

We know: 
3a = d
Since b = d, 
b = 3a

We know:
a = 2e
e = a ÷ 2

Using Eq 4, b = d
8a + 2d = 3d + f
8a = d + f

Using Eq 1, 3a = d
8a – f = 3a
8a – 3a = f
f = 5a

Using Eq 4, c = f
c = 5a

So we have the following equations: 
b = 3a
d = 3a
f = 5a
c = 5a
e = a ÷ 2

Assume, a = 1. We would get:
b = 3 (1) = 3
d = 3 (1) = 3
f = 5 (1) = 5
c = 5 (1) = 5
e = ½

We can let it remain like this. 
However, if we need to express ‘e’ as a whole number, we can simply multiply it by 2. However, we would need to multiply all the other variables by 2 too. 

a = 2 (1) = 2
b = 2 (3) = 6
c = 2 (5) = 10
d = 2 (3) = 6
e = 2 (½) = 1
f = 2 (5) = 10

We would get the following equation: 
2Ca₃(PO₄)₂ + 6SiO₂ + 10C  →  6CaSiO₃ + 1P₄ + 10CO

We can check again to see if everything is balanced. 

Calcium
Number of atoms on LHS = 6
Number of atoms on RHS = 6

Phosphorus
Number of atoms on LHS = 4
Number of atoms on RHS = 4

Oxygen (present in PO₄ nad SiO₂)
Number of atoms on LHS = 16 + 12 = 28
Number of atoms on RHS = 18 + 10 = 28

Silicon
Number of atoms on LHS = 6
Number of atoms on RHS = 6

Carbon
Number of atoms on LHS = 10
Number of atoms on RHS = 10