Question

Balance the following chemical reactions:
(1)$Mn{{O}_{2}}+HCl\to MnC{{l}_{2}}+{{H}_{2}}O+C{{l}_{2}}$
(2) ${{C}_{2}}{{H}_{4}}+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$

Answer 1

The balancing of the equation is the application of conservation of mass. The mass of the reactants side and product side should be the same, i.e. the mass should be conserved, it can neither be created nor be destroyed it can only be interchanged.

Complete step-by-step answer: In the question, two chemical equations of the reactions are given and we have to balance the equation.
Balancing the chemical equation is the process of equalizing the number of atoms on the reactant side and product side. A balanced chemical equation refers to the equation in which the number of atoms of the same kind will be equal on both sides i.e. left side (reactant side) and right side (product side).
And we are familiar that the chemical equation is the simple representation of the reaction taking place and from a balanced equation we could calculate the number of moles of each species taking part in the reaction.
Now let us discuss the steps involved in balancing the chemical equation by solving each problem.
(1) $Mn{{O}_{2}}+HCl\to MnC{{l}_{2}}+{{H}_{2}}O+C{{l}_{2}}$
Now let's write each number of atoms in right and left side of the equation

Reactant side	Product side
Mn	1
O	2
H	1
Cl	1

If we give a stoichiometric value of 4 To HCl, the count of all the atoms in the reactant side and product side will become the same.
So the balanced equation will be: $Mn{{O}_{2}}+4HCl\to MnC{{l}_{2}}+2{{H}_{2}}O+C{{l}_{2}}$
Now check the number of atoms on both sides of the equation.
Now let us discuss the second equation: ${{C}_{2}}{{H}_{4}}+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$
We can solve the question as we did in the above case, first tabulate the number of atoms on each side of the chemical equation.

No. of atoms	Reactant side	Product side
C	2	1
H	4	2
O	2	3

First, let us start balancing the equation from the right side, there are two carbon atoms on the left side but only one on the right side, so multiply with 2.And there are 4 H atoms on the left side but only two on the right side so multiply the left side with 2.
Then the equation becomes:
${{C}_{2}}{{H}_{4}}+{{O}_{2}}\to 2C{{O}_{2}}+2{{H}_{2}}O$
Now we have to balance the O atoms, there are 2 O atoms on the left side but six O atoms on the left side so multiply the O atom of the right side with 3.
The equation will become:
${{C}_{2}}{{H}_{4}}+3{{O}_{2}}\to 2C{{O}_{2}}+2{{H}_{2}}O$
And hence the equation is balanced now.

Note: While balancing the equation only the stoichiometric numbers associated with the atoms only can be changed. The subscript of the atoms should not be changed to balance the equation as it changes the chemical ratio of the compound. While balancing the equation if a group of atoms present inside the parentheses and if it possesses a subscript then that subscript is applicable for all the atoms inside the parentheses.