Question

Skeleton equation of a chemical reaction is $\text{C}{{\text{u}}_{\text{2}}}\text{S + CuS}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ Cu + S}{{\text{O}}_{\text{2}}}$. Write its balanced chemical equation.

Answer 1

Hint: To balance a chemical equation, we need to know the meaning of it at first. By balancing chemical equations, we mean making sure that the number of the different atoms of elements in the reactants side is equal to the side of the products.

Step by step answer:
To begin the balancing of the chemical equation, we need to follow certain steps. Here they are listed one by one:
First, we need to write formulas for the reactants and the products that are provided in the question. The products and the reactants that are given is:
$\text{C}{{\text{u}}_{\text{2}}}\text{S + CuS}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ Cu + S}{{\text{O}}_{\text{2}}}$
Secondly, we are required to count the number of atoms of each element present on the reactant and the product side, so that we can equalise them.
As we can see that the number of atoms of each element is not equal on both sides (that is reactant and product) we need to multiply them with numbers which can help us in making the atoms on both sides of the chemical reaction equal.
So, now it is time for trial and error. Chemical equation balancing is a process which needs to be on the basis of hit and trial. So, unless we find the correct number to multiple the compounds, we need to go on with the process.
As we can see that the number of Cu atoms on the reactant’s side is 3 and on the product’s side is 1, we need to multiply the Cu atom with 3 on the product’s side. The number of atoms of O on the reactant's side is 4 and on the product's side is 2. So we need 2 more O atoms on the product’s side. And lastly, the number of S atoms on the reactant's side is 2 and that of the product's side is 1. So, we need more S atoms on the product’s side.
Let us begin the multiplication. Making the number of Cu atoms equal by multiplying 3 on the product’s side.
$\text{C}{{\text{u}}_{\text{2}}}\text{S + CuS}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ 3Cu + S}{{\text{O}}_{\text{2}}}$
Now, making the S and O equal on both sides. For that we are multiplying 2 with $\text{S}{{\text{O}}_{\text{2}}}$. Now the equation becomes as follows:
$\text{C}{{\text{u}}_{\text{2}}}\text{S + CuS}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ 3Cu + 2S}{{\text{O}}_{\text{2}}}$
In the above equation, all the atoms are equal on both the sides of the reaction. Therefore, we can say that the final balanced equation is:
$\text{C}{{\text{u}}_{\text{2}}}\text{S + CuS}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ 3Cu + 2S}{{\text{O}}_{\text{2}}}$

Note: The main aim of chemical equation balancing is to preserve the law of conservation of mass. The law of conservation of law states that matter can neither be created nor can be destroyed, in chemical reactions. So, the number of atoms of each element, on both the sides of the reactions will be the same.