Question

How can you balance a chemical equation by the hit and trial method? Explain with an example.

Answer 1

Steps involved in balancing a chemical equation are as follows:

• Write the chemical equation in the form of a word equation. Keep the reactants on the left side and the products on the right side. Separate them by an arrow$\left(\to \right)$, whose head points from the reactants towards the products.
• Convert the word equation into the symbol equation by writing the symbols and formulae of all the reactants and products.
• Make the atoms of different elements on both sides of the equation equal by a suitable method. This is known as the balancing of equations.
• Do not change the formulae of the substances while balancing the equation.
• Make the equation more informative if possible.

Example showing balancing of an equation:

Iron reacts with water (steam) to form iron (II, III) oxide and liberates hydrogen gas.

Step 1: Word equation for the reaction

$\mathrm{Iron}+\mathrm{Water}\to \mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}+\mathrm{Hydrogen}$

Step 2: Symbol equation for the same reaction

$\mathrm{Fe}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Fe}}_3{\mathrm{O}}_4+{\mathrm{H}}_2$

Step 3: Balancing of the equations is done in the following steps

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

$\underset{\mathrm{Iron}}{\mathrm{Fe}}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2}$

Element	Reactant side	Product side
Fe	1	3
H	2	2
O	1	4

Thus, the number of H atoms is equal on both sides. At the same time, the number of ‘Fe’ and ‘O’ atoms is not equal.

(ii) To balance the oxygen atoms, put coefficient 4 before $\left({\mathrm{H}}_2\mathrm{O}\right)$ on the reactant side.

$\underset{\mathrm{Iron}}{\mathrm{Fe}}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2}$

(iii) In order to equate Hydrogen atoms, put coefficient 4 before hydrogen on the product side.

$\underset{\mathrm{Iron}}{\mathrm{Fe}}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4}+\underset{\mathrm{Hydrogen}}{4{\mathrm{H}}_2}$

(iv) In order to balance the Fe atoms, put coefficient 3 before Fe on the reactant side.

$\underset{\mathrm{Iron}}{3\mathrm{Fe}}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4}+\underset{\mathrm{Hydrogen}}{4{\mathrm{H}}_2}$

(v) On final inspection, the number of atoms of all the elements on both sides of the equation is equal. Therefore, the equation is balanced.

$\underset{\mathrm{Iron}}{3\mathrm{Fe}}+\underset{\mathrm{Water}}{4{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Iron}(\mathrm{II},\mathrm{III})\mathrm{oxide}}{{\mathrm{Fe}}_3{\mathrm{O}}_4}+\underset{\mathrm{Hydrogen}}{4{\mathrm{H}}_2}$