Question

How can I balance this chemical equation $Pb{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+{{K}_{2}}Cr{{O}_{4}}\text{ }\to PbCr{{O}_{4}}\text{ }+\text{ }KN{{O}_{3}}$ ?

Answer 1

We have to calculate the number of each atom involved in the reactants side and how many numbers of the same chemical are present on the product side and later we have to make them balance by using integer numbers.

Complete step by step answer:
- In the question it is asked to balance the given chemical reaction.
- The given chemical reaction is as follows.
$Pb{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+{{K}_{2}}Cr{{O}_{4}}\text{ }\to PbCr{{O}_{4}}\text{ }+\text{ }KN{{O}_{3}}$
- The above given chemical reaction is an example of double displacement reaction in which two cations are displaced by two anions.
- From the chemical reaction the number of reactants involved in the chemical reaction is
Pb =1, $N{{O}_{3}}$ = 2, K = 2 and $Cr{{O}_{4}}$ = 1.
- The number of products appeared in the given chemical reaction are
Pb = 1, $N{{O}_{3}}$ = 1, K = 1 and $Cr{{O}_{4}}$ = 1.
- We can say that potassium (K) and Nitrate ($N{{O}_{3}}$) are imbalanced in the given chemical reaction.
- We have to multiply the potassium nitrate ($KN{{O}_{3}}$ ) on the product side to get the balanced chemical reaction and it is as follows.
$Pb{{\left( N{{O}_{3}} \right)}_{2}}\text{ }+{{K}_{2}}Cr{{O}_{4}}\text{ }\to PbCr{{O}_{4}}\text{ }+\text{ 2}KN{{O}_{3}}$
- In the above chemical reaction one mole of lead nitrate reacts with one mole of potassium chromate and forms one moles of lead chromate and two moles of potassium nitrate as the products.

Note: We have to leave the polyatomic ions like $N{{O}_{3}}$ and $Cr{{O}_{4}}$ together when they are together on the both sides of the chemical reactions. We have to count them as one unit, not separate elements while counting.