Question

The balanced equation for the following chemical reactions. Anhydrous potassium nitrate is heated with an excess of metallic potassium.
$KN{{O}_{3}}(s)+K(s)\to$
(A)- $2KN{{O}_{3}}(s)+10K(s)\to 6{{K}_{2}}O+{{N}_{2}}$
(B)- $2KN{{O}_{3}}(s)+10K(s)\to 6{{K}_{2}}O+2N{{H}_{3}}$
(C)- $2KN{{O}_{3}}(s)+10K(s)\to 6{{K}_{2}}O+2N{{H}_{3}}+2KOH$
(D)- $2KN{{O}_{3}}(s)+10K(s)\to 6{{K}_{2}}O+N{{H}_{3}}+KOH$

Answer 1

Reaction of anhydrous potassium nitrate ($KN{{O}_{3}}$) with potassium (K) metal one of the methods of preparation of potassium oxide (${{K}_{2}}O$).
According to the law of conservation of mass, a balanced chemical equation must have the same number of atoms of each element on the reactant and product sides.

Complete step by step solution:
Potassium oxide (${{K}_{2}}O$) is a pale yellow coloured ionic compound. It is prepared by heating anhydrous potassium nitrate ($KN{{O}_{3}}$) with excess potassium metal. Nitrogen gas (${{N}_{2}}$) is evolved along with the formation of (${{K}_{2}}O$).

Now let us look for the chemical equation representing the above reaction between $KN{{O}_{3}}$ and K among the given options.
Option (A)- $2KN{{O}_{3}}(s)+10K(s)\to 6{{K}_{2}}O+{{N}_{2}}$
One molecule of ${{K}_{2}}O$ has 2 K atoms, so there are a total 12 K atoms on the product side. 10 K-atoms from potassium metal and 2 atoms from $KN{{O}_{3}}$ makes a total of 12 K-atoms on the reactant side.
The number of O-atoms are equal, i.e. 6 on the reactant and product side. There are 2 N-atoms on both sides of the equations.
Therefore, the above equation represents the balanced chemical equation for the given reaction.
There is no formation of ammonia ($N{{H}_{3}}$) or potassium hydroxide (KOH) during the preparation of ${{K}_{2}}O$, therefore, none of the other options can be correct.

Hence, the correct option is (A).

Note: We can answer the question even if we do not know the products of the reaction. There is no H atom on the reactant side in the given reaction so the balanced chemical equation must also not have any H atom on the product side. All the three options except for option (A) have H-atoms on the reactant side, and hence they are incorrect.