Question

Define the hit and trial method for balancing the chemical equation.

Answer 1

Balancing of chemical equations:

• The process of making an equal number of different types of elements, on both sides of the equation, is known as balancing the chemical equation.
• The balancing of a chemical equation is done with the help of the hit and trial method.
• In this method, the coefficients before the symbols or formulae of the reactants and products are adjusted in such a way that the total number of atoms of each element on both sides of the arrowhead becomes equal.
• This balancing is also known as mass balancing because if the atoms of elements on both sides are equal and their masses will also be equal.
• Example: Iron reacts with water (steam) to form iron (II, III) oxide and liberates hydrogen gas.
• The word equation for the reaction is$\mathrm{Iron}+\mathrm{Water}\to \mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}+\mathrm{Hydrogen}$
• The symbol equation for the same reaction is $\mathrm{Fe}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{H}}_2$

The balancing of the equations is done in the following steps:

Let us count the number of atoms of all the elements in the reactants and products on both sides of the equation.

Element	No. of atoms of reactants (L.H.S.)	No. of atoms of products(R.H.S.)
$\mathrm{Fe}$	1	3
$\mathrm{H}$	2	2
$\mathrm{O}$	1	4

Hit and trial method:

• To check the number of atoms on both sides and make them equal by using the numerals.
• Thus, the number of $\mathrm{H}$ atoms is equal on both sides. At the same time, the number of $\mathrm{Fe}$ and $\mathrm{O}$ atoms is not equal.
• On inspection, the number of $\mathrm{O}$ atoms in the reactant (${\mathrm{H}}_2\mathrm{O}$) is 1 while in the product (${\mathrm{Fe}}_3{\mathrm{O}}_4$), these are 4. To balance the atoms, put coefficient 4 before (${\mathrm{H}}_2\mathrm{O}$) on the reactant side. The partially balanced equation may be written as
  $\mathrm{Fe}+4{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{H}}_2$
• In order to equate H atoms, put coefficient 4 before ${\mathrm{H}}_2$ on the product side. As a result, the H atoms on both sides of the equation become 8 and are thus balanced. The partially balanced equation may now be written as
  $\mathrm{Fe}+4{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2$
  In order to balance the Fe atoms, put coefficient 3 before $\mathrm{Fe}$ on the reactant side. The equation formed may be written as -
  $3\mathrm{Fe}+4{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2$
• On final inspection, the number of atoms of all the elements on both sides of the equation is equal. Therefore, the equation is balanced.
• $\underset{\mathrm{Iron}}{3\mathrm{Fe}\left(\mathrm{s}\right)}+\underset{\mathrm{Steam}}{4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{Oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{4{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Element	No. of atoms of reactants (L.H.S.)	No. of atoms of products(R.H.S.)
$\mathrm{Fe}$	3	3
$\mathrm{H}$	8	8
$\mathrm{O}$	4	4

Hence, the final equation has equal numbers of atoms on both sides. It is a well-balanced equation.