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 +

**a**O₂ → 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.

**a**Ca₃(PO₄)₂ +** b**SiO₂ + **c**C → **d**CaSiO₃ + **e**P₄ + **f**CO

**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:

**2**Ca₃(PO₄)₂ +** 6**SiO₂ + **10**C → **6**CaSiO₃ + **1**P₄ + **10**CO

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