When the number of atoms of each element on both sides of the arrow in a chemical equation is equal, it is called **balanced chemical equation**. This is because no matter is created or destroyed during a chemical reaction according to the law of conservation of mass. Therefore, we must balance the chemical equation to satisfy the requirement of the law of conservation of mass.

The way by which the number of atoms of each element on both sides of the arrow in a chemical equation is made equal is called **balancing of chemical equation**. We can do balancing a chemical equation by any of the following method:

- Hit and trail method
- Partial equation method
- Oxidation number method
- Ion-electron method

In this tutorial, we will understand the first two methods. The other two methods will understand in the further tutorial.

## Hit and Trial Method to Balance Chemical Equation

Hit and trial method can balance many chemical equations. This method is also called inspection method or trial and error method. Balancing a chemical equation using this method has the following steps:

**Step 1:** Write down the skeleton equation by writing the correct formulas of reactants and products.

**Step 2:** If any elementary gas (e.g. H2, O2, N2, etc.) or diatomic is present on either side of the chemical equation, write it in the atomic state in the skeleton equation.

**Step 3:** Start to balance from the formula that contains the maximum number of atoms. In case you do not find it convenient, start balancing atoms which are present the least number of times.

**Step 4:** Balance the number of atoms present in the elementary gas or diatomic in the last.

**Step 5:** If the balancing is done, convert the equation into the molecular form.

It is important to note that an integer value placed in the front of a chemical formula multiplies each atom of that formula by that integer value. For example:

- 2NH
_{3}means two nitrogen atoms and 6 hydrogen atoms - 3H
_{2}O means 6 hydrogen atoms and 3 oxygen atoms

Let’s take some examples to balance chemical equations by this method.

## Application of Hit and Trail Method with Examples

**Example 1: Balance the given unbalanced chemical equation**:

CH_{4}(g) + O_{2}(g) → CO_{2}(g) + H_{2}O(l)

**Solution:**

**Step 1:** The skeleton equation is as: CH_{4}(g) + O(g) → CO_{2}(g) + H_{2}O(l)

In this chemical equation, elementary gas equation is in the atomic form.

**Step 2:** Since CH_{4} molecule in the above chemical equation contains the maximum number of atoms, so we will start to balance from here.

(a) As you can see that number of carbon atoms on both sides are equal.

(b) In the reactant side, CH_{4} molecule contains four H atoms, while there are only 2 H atoms in H_{2}O in the product side.

So, we can balance H atoms by multiplying H2O by 2. In this way, we can write partially balanced equation as:

CH_{4}(g) + O(g) → CO_{2}(g) + 2H_{2}O(l)

**Step 3:** Now C atoms and H atoms are balanced in the above chemical equation. We will balance oxygen atoms by multiplying O(g) on the left by 4 because there are four oxygen atoms in the product side. Thus, the chemical equation we can write as:

CH_{4}(g) + 4O(g) → CO_{2}(g) + 2H_{2}O(l)

Now this chemical equation is balanced because the number of atoms of each element on either side are equal.

**Step 4:** At last, convert this equation into molecular form. We can write for it as:

CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(l)

**Example 2: **

When acetylene burns in oxygen, it forms carbon dioxide and water. Write the skeleton equation for the chemical reaction and balance the chemical equation using hit and trial method.

**Solution:**

**Step 1:** The skeleton equation is as:

C_{2}H_{2}(g) + O(g) → H_{2}O(l) + CO_{2}(g)

Here, we have written the elementary gas oxygen in the atomic form.

**Step 2:** C_{2}H_{2}(g) has the maximum number of atoms. So, we will start to balance from here. With skeleton equation, H-atoms on both sides are equal. We can balance C-atoms by multiplying CO_{2}(g) on the right side by 2. So, the partially balanced equation is:

C_{2}H_{2}(g) + O(g) → H_{2}O(l) + 2CO_{2}(g)

**Step 3:** Now, there are five O-atoms on the product side and only one on the reactant side. Therefore, we need to balance it. To balance the number of O-atoms on both sides, we will multiplying it on the reactant side by 5. The resulting equation is:

C_{2}H_{2}(g) + 5O(g) → H_{2}O(l) + 2CO_{2}(g)

**Step 4:** This chemical equation is balanced. Now convert this atomic equation into molecular form by multiplying throughout by 2. Thus,

2C_{2}H_{2}(g) + 5O_{2}(g) → 2H_{2}O(l) + 4CO_{2}(g)

This is the resulting balanced equation.

**Example 3: **Balance the following unbalanced chemical equations:

**1. KMnO _{4} + HCl → KCl + MnCl_{2} + H_{2}O + Cl_{2}**

**Solution:**

**Step 1:** The skeleton equation is:

KMnO_{4} + HCl → KCl + MnCl_{2} + H_{2}O + Cl (Writing elementary gas in atomic form)

**Step 2:** Starting KMnO_{4}, K and Mn are balanced on both sides.

**Step 3:** Balance O atoms on both sides:

KMnO_{4} + HCl → KCl + MnCl_{2} + 4H_{2}O + Cl

**Step 4:** Balance H atoms on both sides:

KMnO_{4} + 8HCl → KCl + MnCl_{2} + 4H_{2}O + Cl

**Step 5:** Balance Cl atoms on both sides:

KMnO_{4} + 8HCl → KCl + MnCl_{2} + 4H_{2}O + 5Cl (This is the balanced atomic equation)

**Step 6:** Making it molecular:

2KMnO_{4} + 16HCl → 2KCl + 2MnCl_{2} + 8H_{2}O + 5Cl_{2}

Thus, the balanced molecular equation is as:

2KMnO_{4}(aq) + 16HCl(aq) → 2KCl(aq) + 2MnCl_{2}(aq)+ 8H_{2}O(l) + 5Cl_{2}(g)

**2. H _{3}PO_{2} → H_{3}PO_{4} + PH_{3}**

**Solution:**

Since there are two P atoms in the product side, therefore, we will multiplying the coefficient of H_{3}PO_{2} by 2. Thus, the equation is:

2H_{3}PO_{2} → H_{3}PO_{4} + PH_{3}

This is the balanced chemical equation.

**3. Ca + H _{2}O → Ca(OH)_{2} + H_{2}**

**Solution:**

**Step 1: **Starting from Ca(OH)_{2} Ca atom is already balanced.

**Step 2:** To balance H atom on both sides, we will have to multiply the coefficient of H_{2}O by 2. Thus, the chemical equation obtained is:

Ca + 2H_{2}O → Ca(OH)_{2} + H_{2}

Now, O atom is also balanced in this process. So, the balanced chemical equation is as:

Ca + 2H_{2}O → Ca(OH)_{2} + H_{2}

## Partial Equation Method to Balance Chemical Equation

Complicated chemical equations which contain many reactants and products we cannot balance easily by using hit and trail method. Such chemical equations are balanced by using partial equation method. There are the following steps to balance a chemical equation by using partial equation method. They are as:

**Step 1:** Write the given chemical equation into two or more different probable steps. A chemical equation written into two or more steps is called partial equations or simpler reactions.

**Step 2:** Balance each partial equation separately by using hit and trial method as discussed earlier.

**Step 3:** Multiply such balanced partial equations with suitable coefficients so as to exactly the common substances (compounds or molecules) which is not present in the overall chemical equation.

**Step 4:** Finally, add the balanced partial chemical equations obtained in step 3. Thus, we will get a balanced chemical equation.

## Application of Partial Equation Method with Examples

**Example 1:** Balance the following chemical equation by using partial equation method.

NaOH + Cl_{2} → NaCl + NaClO_{3} + H_{2}O

**Solution:**

**Step 1:** Split the skeleton equation, NaOH + Cl_{2} → NaCl + NaClO_{3} + H_{2}O into two partial equations as:

Partial eq1: NaOH + Cl_{2} → NaCl + NaClO +H_{2}O

Partial eq2: NaClO → NaClO_{3} + NaCl

**Step 2:** Balance the above two partial equations using hit and trail method.

Balanced partial eq1: 2NaOH + Cl_{2} → NaCl + NaClO + H_{2}O

Balanced partial eq2: 3NaClO → NaClO_{3} + 2NaCl

**Step 3: **As NaClO is not present in the overall equation, so to cancel it exactly, we will multiply the balanced partial eq1 by 3 and consider it balanced partial eq3. Then,

2NaOH + Cl_{2} → NaCl + NaClO + H_{2}O ] * 3

Balanced partial eq3: 6NaOH + 3Cl_{2} → 3NaCl + 3NaClO + 3H_{2}O

**Step 4:** Now add the balanced partial eq3 and balanced partial eq2 on both sides. Thus,

6NaOH + 3Cl_{2} + 3NaClO → 3NaCl + 3NaClO + 3H_{2}O + NaClO3 + 2NaCl

6NaOH + 3Cl_{2} → 5NaCl + NaClO_{3} + 3H_{2}O (This is the required balanced chemical equation)

**Example 2:** In an acidic solution, KMnO_{4} oxidizes ferrous sulphate (FeSO_{4}) to ferric sulphate (Fe_{2}(SO_{4})_{3}). Write the skeleton equation for this reaction and balance it using partial equation method.

**Solution:**

**Step 1:** The skeleton equation is:

KMnO_{4} + H_{2}SO_{4} + FeSO_{4} → K_{2}SO_{4} + MnSO_{4} + Fe_{2}(SO_{4})_{3} + H_{2}O

**Step 2:** Split this skeleton equation into two partial equations as:

Partial eq1: KMnO_{4} + H_{2}SO_{4} → K_{2}SO_{4} + MnSO_{4} + H_{2}O + O

Partial eq2: FeSO_{4} + H_{2}SO_{4} + O → Fe_{2}(SO_{4})_{3} + H_{2}O

**Step 3:** Now balance the above both partial equations as:

Balanced partial eq1: 2KMnO_{4} + 3H_{2}SO_{4} → K_{2}SO_{4} + 2MnSO_{4} + 3H_{2}O + 5O

Balanced partial eq2: 2FeSO_{4} + H_{2}SO_{4} + O → Fe_{2}(SO_{4})_{3} + H_{2}O

**Step 4: **To cancel out the common element, multiply the balanced partial eq2 by 5 and consider it as a third balanced partial equation.

2FeSO_{4} + H_{2}SO_{4} + O → Fe_{2}(SO_{4})_{3} + H_{2}O ] * 5

Balanced partial eq3: 10FeSO_{4} + 5H_{2}SO_{4} + 5O → 5Fe_{2}(SO_{4})_{3} + 5H_{2}O

**Step 5:** Add balanced partial eq1 and balanced partial eq3 on both sides. Then, we will get,

2KMnO_{4} + 3H_{2}SO_{4} + 10FeSO_{4} + 5H_{2}SO_{4} + 5O → K_{2}SO_{4} + 2MnSO_{4} + 3H_{2}O + 5O + 5Fe_{2}(SO_{4})_{3} + 5H_{2}O

The overall balanced equation is as:

2KMnO_{4} + 8H_{2}SO_{4} + 10FeSO_{4} → K_{2}SO_{4} + 2MnSO_{4} + 5Fe_{2}(SO_{4})_{3} + 8H_{2}O

In this chapter, you have learned how to balance chemical equation with the help of important examples. Hope that you will have understood the steps discussed explaining the balancing chemical equation and practiced all examples.

Thanks for reading!!!