# What is balancing chemical equations?

Balancing chemical equations is a mathematical operation performed to maintain the law of conservation of matter in reactions. Matter is neither created nor destroyed. Therefore, the number of atoms before and after a chemical reaction must be the same, both in the reactants and in the products.

To balance a chemical equation you have to count the atoms before and after the arrow. If the numbers match, it means that the reaction is balanced. Otherwise, it will be necessary to equal these numbers by modifying the stoichiometric coefficients through trial and error, this being the simplest and most intuitive method for balancing.

When swinging by trial and error it is as if atoms were added or removed on one of the two sides of the seesaw. Until the seesaw is suspended horizontally, the chemical equation will not be fully balanced. There can be no more (create matter) and no fewer atoms (destroy matter) on either side.

__Easy Equation Balancing Examples__

__Easy Equation Balancing Examples__

Balanced or easy-to-balance chemical equations will be seen in the following examples. All must have in common that the number of atoms, for each of the elements, will be the same before and after the arrow once the chemical equation has been balanced.

**Mercury oxidation**

2Hg + O _{2} → Hg _{2} O _{2}

We have two elements: mercury (Hg) and oxygen (O). We express the number of its atoms before (reactants) and after (products) of the arrow:

Hg: 2 (left) – 2 (right)

O: 2 (left) – 2 (right)

The equation is balanced because the Hg and O atoms are the same on both sides of the arrow.

**Hydrogen combustion**

H _{2} + O _{2} → H _{2} O

We again have two elements: hydrogen (H) and oxygen (O). Repeating the previous step:

H: 2 – 2

O: 2 – 1

There is one more oxygen left over on the left side than on the right. We must therefore add another oxygen to the right. To do this, we modify the stoichiometric coefficients, which are the numbers that accompany and precede the formulas. If we place a 2 in front of H _{2} O it will give us:

H _{2} + O _{2} → 2H _{2} O

H: 2 – 4

O: 2 – 2

The oxygens are balanced. But not the hydrogens. You have to add two hydrogens now on the left:

2H _{2} + O _{2} → 2H _{2} O

H: 4 – 4

O: 2 – 2

And the equation is finally balanced: same number of H and O before and after the arrow.

Note that the stoichiometric coefficients (2 for H _{2} and 1 for O _{2} ) multiply the subscripts of the formulas. This multiplication will give us the total number of atoms for a given element.

**Magnesium reduction**

Mg (NO _{3} ) _{2} + 2Li → Mg + 2LiNO _{3}

We evaluate the numbers of atoms for each element:

Mg: 1 – 1

Li: 2 – 2

N: 2 – 2

Or: 6 – 6

The chemical equation is already balanced and there is no need to balance it.

__Solved exercises__

__Solved exercises__

Before proceeding to balance any equation, it is always necessary to confirm if it is not already balanced. Having said that, we proceed with the following exercise:

**Exercise 1**

**Balanced or not?**

Fe _{2} O _{3} + H _{2} O → Fe (OH) _{3}

We have three elements: iron (Fe), hydrogen and oxygen. One suggestion when balancing equations is to start by counting the numbers of atoms of the least abundant element in the equation; which is generally different from H and O. Thus, we first count the Fe atoms:

Faith: 2 – 1

The subscript 3 of (OH) _{3} does not multiply the Fe on the left. For H and O we have:

H: 2 – 3

Or: 4 – 3

All the elements are out of balance.

**Swinging**

Before even balancing the H and O, we must balance the Fe atoms: the least abundant element, since there is hardly 3 Fe, unlike the 5 H and 7 O. We must add, then, one Fe to the right:

Faith: 2 – 2

Which is the same as putting a 2 as a stoichiometric coefficient in front of Fe (OH) _{3} . The equation will be as:

Fe _{2} O _{3} + H _{2} O → 2Fe (OH) _{3}

Faith: 2 – 2

H: 2 – 6

Or: 4 – 6

But the H’s and O’s remain unbalanced. An H _{2} O contributes 2 H and we are missing 4 H. Therefore, we will add another two H _{2} O so that we have three H _{2} O:

Fe _{2} O _{3} + 3H _{2} O → 2Fe (OH) _{3}

And we evaluate again:

Faith: 2 – 2

H: 6 – 6

Or: 6 – 6

Notice how the O’s balanced on their own once we balanced the H’s. The equation is finally balanced.

**Exercise 2**

**Balanced or not?**

CH _{4} + O _{2} → CO _{2} + H _{2} O

We evaluate the number of atoms for all the elements present:

C: 1 – 1

H: 4 – 2

Or: 2 – 3

Carbon is balanced, but the same is not true for hydrogen and oxygen. The equation is out of balance.

**Swinging**

As C is balanced, we do not want to modify the stoichiometric coefficients of CH _{4} or CO _{2} . At least not in principle.

We focus our attention on H and O, especially H. Why? Because looking at the equation it is easier to balance H before O. We therefore need to add 2 H to the right, placing a coefficient 2 in front of H _{2} O:

CH _{4} + O _{2} → CO _{2} + 2H _{2} O

C: 1 – 1

H: 4 – 4

O: 2 – 4

C and H are balanced. We have 2 O left. If each O _{2} contributes 2 O, then we will add an extra O _{2} to make it 4 O to the left. This is the same as placing a coefficient 2 in front of O _{2} :

CH _{4} + 2O _{2} → CO _{2} + 2H _{2} O

C: 1 – 1

H: 4 – 4

O: 4 – 4

And the equation is finally balanced.

**Exercise 3**

**Balanced or not?**

N _{2} + H _{2} → NH _{3}

We evaluate for our two elements N and H:

N: 2 – 1

H: 2 – 3

The equation is not balanced.

**Swinging**

We start with a balanced N, the element other than H. We lack one N on the right, and as each NH _{3} contributes 1 N, we must add another NH _{3} to have 2 N:

N _{2} + H _{2} → 2NH _{3}

N: 2 – 2

H: 2 – 6

And now we are missing H on the left. If each H _{2} contributes 2 H to the equation, then we must add another two so that there are 6 H. This is the same as placing a stoichiometric coefficient 3 in front of H _{2} :

N _{2} + 3H _{2} → 2NH _{3}

N: 2 – 2

H: 6 – 6

And the chemical equation is finally balanced.