How to Balance Chemical Equations by hit and trial method

Today, we will discuss how to balance chemical equations by Hit and Trial method. As a class 10^th student, you must have gone through chapter 1 chemical reactions and equations. In this chapter, you must have read about chemical reactions.

You know a chemical reaction is written as a ‘chemical equation’. In this lesson, we will learn how to balance chemical equation step by step. But before we go ahead you should know some important things.

Why should chemical equations be balanced?

This is based on the law ‘Conservation of Mass’. According to this law-

mass can neither be created nor destroyed in a chemical reaction.

The total mass of elements present in reactants must be equal to the total mass of elements present in products.

So the number of atoms of elements on the reactant side of the equation should be equal to the number of atoms of the same elements on the product side.

That’s why a chemical equation should be balanced.

What is a chemical reaction?

A chemical change in the properties (i.e.- composition) of a substance is called a chemical reaction.

What is a chemical equation?

The representation of a chemical reaction by using symbols and formulas of reactants and products is called a chemical equation.

What are the types of chemical equations?

A chemical equation is of the following types-

i.              Word equation

ii.          Skeletal equation

iii.       Balance equation

i.              Word equation - When a chemical reaction is described in words, the equation is called a ‘word equation’.

Magnesium + oxygen → magnesium oxide

In this equation, only the names of reactants and products are written.

 

ii.          Skeletal equation – When symbols or formulas of reactants and products are written and the mass of substances on both sides is not the same, the equation is called a skeletal equation.

                Mg + O₂   → MgO

iii.       Balanced equation- The chemical equation in which the number of atoms of elements is equal in reactant and product side is called a balanced equation.

                               2Mg + O₂   → 2MgO

 

Now, we can easily understand different types of chemical equations.

Steps in balancing Chemical Equation

To find whether the chemical equation is balanced or not, we will use the ‘Hit and Trail Method’ that has been discussed in class 10 Science.

We will learn how to balance a chemical equation through a simple example.

Let’s say – hydrogen reacts with oxygen to produce water.

  Hydrogen + Oxygen → Water (Word Equation)

        H₂ + O₂ → H₂O (unbalanced equation)

Note: - Reactants are on the left-hand side and products are on the right-hand side.

Step -1

Firstly, draw a box around each reactant and product so that we can’t change any number in the box.

Step-2

 

Write the number of atoms of each element on the reactants and product sides.

Elements	Number of atoms on the reactant side	Number of atoms on the product side
Hydrogen	2	2
Oxygen	2	1

 

Step-3

Find the atom with the maximum number in the reactant or product side and try to balance it.

Note: We cannot change the formula of any compound.

In the given reaction, we can identify that oxygen has 2 atoms on the reactant side and 1 atom on the product side, so we put coefficient ‘2’ before water (H₂O) on the product side to balance the oxygen atom.                  

                     H₂ + O₂ → 2H₂O

Step -4

Now check the whole equation and count the number of atoms on both sides.

 

Elements	Number of atoms in the reactant side	Number of atoms on the product side
Hydrogen	2	2×2 =4
Oxygen	2	2×1=2

 

We find oxygen atom is balanced on both sides but the hydrogen atom has become unbalanced. so we will balance the hydrogen atom.

Step-5

Now put coefficient ‘2’ before the hydrogen atom at the reactant side.

                     2H₂ + O₂ → 2H₂O

 

Now again check the whole equation and count the number of atoms on both sides.

Elements	Number of atoms on the reactant side	Number of atoms on the product side
Hydrogen	2×2=4	2×2 =4
Oxygen	2	2×1=2

 

We can see on the reactant side 4 hydrogens and 2 oxygen and on the product side 4 hydrogens and 2 oxygen atoms, so the reaction is balanced.

Step-6

When the given equation got balanced then we will write symbols for the physical state (solid, liquid, or gas).

Sometimes the reaction conditions like pressure, temperature, catalyst, etc are also written.

                     2H_2(g) + O_2(g) → 2H₂O_(l)

.........................................................................................

Now we will take another example

Hydrogen sulphide gas burns in the air to give water and sulphur dioxide.

Hydrogen + oxygen→ water + sulphur dioxide

(A substance reacts with oxygen when it starts burning.)

             H₂S + O₂ → H₂O + SO₂

    1. Draw a box around all the substances

    2. Count the number of atoms of reactants and products

 

Elements	Number of atoms in reactant side	Number of atoms in product side
Hydrogen	2	2
Sulphur	1	1
Oxygen	2	3

 

We can see oxygen has 2 atoms on the reactant side and 3 atoms on the product side.

So we will balance oxygen by the ‘Hit and Trail method’.

    3. We put 2 before oxygen on the reactant side.

                H₂S + 2O₂ → H₂O + SO₂

Now again we will count atoms on both sides.

Elements	Number of atoms on reactant side	Number of atoms on product side
Hydrogen	2	2
Sulphur	1	1
Oxygen	4	3

 

 

    4. Now we try to balance ‘oxygen’ and put 2 before H₂O.

                H₂S + 2O₂ → 2H₂O + SO₂

Count atoms on both sides again.

Elements	Number of atoms on reactant side	Number of atoms on product side
Hydrogen	2	2×2=4
Sulphur	1	1
Oxygen	4	4

 

We find hydrogen has 4 atoms on the product side and 2 atoms on the reactant side.

   5. So we will balance hydrogen and put 2 before H₂S.

              2H₂S + 2O₂ → 2H₂O + SO₂

Count atoms on both sides again.

Elements	Number of atoms on reactant side	Number of atoms on product side
Hydrogen	2×2=4	2×2=4
Sulphur	1×2=2	1
Oxygen	4	4

 

We find sulphur has 2 atoms on the reactant side and 1 atom on the product side.

    6. So we will balance sulphur and put 2 before SO₂.

        2H₂S + 2O₂ → 2H₂O + 2SO₂

Count atoms on both sides again.

Elements	Number of atoms on reactant side	Number of atoms on product side
Hydrogen	2×2=4	2×2=4
Sulphur	1×2=2	1×2=2
Oxygen	4	6

 

But again oxygen got imbalanced.

 

    7. Now we will balance oxygen again and remove 2 and put 3 before O₂.

          2H₂S + 3O₂ → 2H₂O + 2SO₂

Count atoms on both sides again.

Elements	Number of atoms on reactant side	Number of atoms on product side
Hydrogen	2×2=4	2×2=4
Sulphur	1×2=2	1×2=2
Oxygen	3×2=6	6

 

    8. Finally, we can see numbers of atoms of all elements in the given reaction are balanced.

 9. Now write different symbols of the physical state of reactants and products.

      2H₂S_(g) + 3O_2(g) → 2H₂O_(i) + 2SO_2(g)

Hope, you understand how to balance chemical equations.

Balance the following equations

1. CH₄ +O₂ → CO₂ + H₂O

2. Fe + O₂  → Fe₂O₃

3. PbO₂+ H₂SO₄→ PbSO₄ + H₂O + O₂

4. Al+ Fe₂O₃ →Al₂O₃ +  Fe

5. KClO₃ → KCl +O₂

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