Question

How do you balance ${\text{PbS}}{{\text{O}}_{\text{4}}}\, \to \,\,{\text{PbS}}{{\text{O}}_3} + \,{{\text{O}}_{\text{2}}}$?

Answer 1

When a chemical reaction is written using the chemical formulae of the compound or species involved in the reaction then it is known as a chemical equation.
When there are the same numbers of atoms of elements on the reactant as well as the product side of the chemical equation then it is a balanced chemical equation.
Steps used to the balanced chemical equation are as follows:
1. Write the given chemical equation.
2. Count all atoms of elements on both the side of the reaction.
3. First balance the element which is present in only one compound on reactant as well as product.
4. Then balance the remaining elements which are found in more than one compound.
5. Count all atoms on the reactants and product side to verify whether all are balanced or not.

Complete step-by-step answer: The stepwise balancing of the given reaction is as follows:
To balance the chemical equation the first step is to write the chemical equation.
${\text{PbS}}{{\text{O}}_{\text{4}}}\, \to \,\,{\text{PbS}}{{\text{O}}_3} + \,{{\text{O}}_{\text{2}}}$

Now, count each type of atom present on both sides of the reaction.
${\text{PbS}}{{\text{O}}_{\text{4}}}\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,{\text{PbS}}{{\text{O}}_3}\,\,\,\, + \,\,\,\,\,{{\text{O}}_{\text{2}}}$
${\text{Pb}}\,\,\,\,\,{\text{1S}}\,\,\,4{\text{O}}\,\,\,\,\,\,\,\,\,\,\,\,1{\text{Pb}}\,\,\,1{\text{S 5O}}$

Now, platinum and sulfur are present in only one species on reactant as well as on the product side. Therefore, balance its number first.
The numbers of platinum also sulfur on reactant as well as on the product side are the same hence, no need to balanced it.
Now, balance oxygen as there is four oxygen on the reactant side but 5 on the product side. Hence, multiply ${\text{PbS}}{{\text{O}}_{\text{4}}}$ and ${\text{PbS}}{{\text{O}}_3}$ by 4 and ${{\text{O}}_{\text{2}}}$ by 2.
${\text{4PbS}}{{\text{O}}_{\text{4}}}\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,4{\text{PbS}}{{\text{O}}_3}\,\,\,\, + \,\,\,\,\,2{{\text{O}}_{\text{2}}}$
${\text{4Pb}}\,\,\,\,\,4{\text{S}}\,\,\,16{\text{O}}\,\,\,\,\,\,\,\,\,\,\,\,4{\text{Pb}}\,\,\,4{\text{S 16O}}$
Thus, oxygen atoms are balanced.
Here, we can see that after balancing oxygen atoms, platinum and sulfur atoms also get balanced.
Now, if we count all the atoms in the equation their number is the same on both sides.
Hence, the chemical equation gets balanced.
${\text{4PbS}}{{\text{O}}_{\text{4}}}\, \to \,\,4{\text{PbS}}{{\text{O}}_3}\, + \,\,2{{\text{O}}_{\text{2}}}$
This is the balanced chemical equation.

Note: When a chemical reaction is written using the chemical formulae of the compound or species involved in the reaction then it is known as a chemical equation.When there are the same numbers of atoms of elements on reactant as well as the product side of the chemical equation then it is a balanced chemical equation.The balanced chemical equation is very important to determine the stoichiometry of the reaction. The stoichiometric ratio obtained between the reactants and products is important to determine the yield of the species formed in the reaction.The stoichiometric ratio is nothing but the mole ratio between the reactants and the products of the reaction.