Question

When $Pb{O_2}$ reacts with conc. $HN{O_3}$ , the gas evolved is:
A. $N{O_2}$
B. ${O_2}$
C. ${N_2}$
D. ${N_2}O$

Answer 1

Hint: We know that in simple terms, it can be said that in chemical reactions, atoms of reactant(s) get rearranged to give the products.

Complete step by step answer:
As per Dalton’s atomic theory, mass is conserved during a chemical reaction and its only reorganization of atoms that gives rise to new products. This reorganization can take place in different ways giving different types of reactions and products.
Here, we will talk about an oxoacid of nitrogen namely nitric acid which has a formula $HN{O_3}$. It has been established that it is a strong acid and gets ionized as per the following reaction:
$HN{O_3} + {H_2}O \to NO_3^ - + {H_3}{O^ + }$
$HN{O_3}$acid is a strong oxidizing agent and can oxidize various metals. However the products would vary as other factors also play significant role such as the concentration of $HN{O_3}$ acid. For example when $Cu$ reacts with dilute $HN{O_3}$ it gives $NO$ gas whereas if concentrated $HN{O_3}$ is used, we get $N{O_2}$ gas. The reactions can be shown by the following chemical equations:
\[\begin{array}{c}
Cu + \;dil.\;HN{O_3} \to Cu{\left( {N{O_3}} \right)_2} + NO + {H_2}O\\
Cu + \;conc.\;HN{O_3} \to Cu{\left( {N{O_3}} \right)_2} + N{O_2} + {H_2}O
\end{array}\]
$HN{O_3}$ acid can also be used to prepare ${N_2}O$ by oxidizing zinc with dilute $HN{O_3}$ acid as follows:
\[Zn + \;dil.\;HN{O_3} \to Zn{\left( {N{O_3}} \right)_2} + {N_2}O + {H_2}O\]
We can prepare ${N_2}{O_5}$ as well by using $HN{O_3}$ acid and ${P_4}{O_{10}}$ as per the following reaction:
$HN{O_3} + {P_4}{O_{10}} \to HP{O_3} + {N_2}{O_5}$
It is evident that $HN{O_3}$ acid is quite significant in synthesis chemistry. Here, we will look at the reaction of concentrated $HN{O_3}$ acid with $Pb{O_2}$. The reaction can be written as follows:
$Pb{O_2} + conc.\;HN{O_3} \to Pb{\left( {N{O_3}} \right)_2} + {O_2} + {H_2}O$
So, we can see that the reaction of nitric acid and lead oxide gives ${O_2}$ gas.

Therefore, from the above explanation the correct option is (B).

Note:
We have to be careful while deducing the products of a reaction when only reactants are given because in addition to the reactants, many other factors are also significant such as reaction conditions or the concentration of the reactants.