Question

Upon heating $\text{KCl}{\text{O}}_3$ in the presence of catalysis amount of $\text{Mn}{\text{O}}_2$, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure $\text{HN}{\text{O}}_3$ gives Y and Z.
(a) W and X are respectively
      A. ${\text{O}}_2$ and ${\text{P}}_4{\text{O}}_{10}$
      B. ${\text{O}}_2$ and ${\text{P}}_4{\text{O}}_6$
      C. ${\text{O}}_3$ and ${\text{P}}_4{\text{O}}_6$
      D. ${\text{O}}_3$ and ${\text{P}}_4{\text{O}}_{10}$
(b) Y and Z are respectively
       A. ${\text{N}}_2{\text{O}}_5$ and $\text{HP}{\text{O}}_3$
       B. ${\text{N}}_2{\text{O}}_4$ and $\text{HP}{\text{O}}_3$
       C. ${\text{N}}_2{\text{O}}_4$ and ${\text{H}}_3\text{P}{\text{O}}_3$
       D. ${\text{N}}_2{\text{O}}_3$ and ${\text{H}}_3\text{P}{\text{O}}_4$

Answer 1

Hint: The given questions can be solved by understanding the nitrogen and oxygen family properly since phosphorus comes under the nitrogen family. $\text{Mn}{\text{O}}_2$ acts as an oxidizing agent and reducing agent.

Complete step by step solution:
The given question can be expressed as given below:
$$\begin{gathered} 3\text{KCl}{\text{O}}_3\underset{\text{Mn}{\text{O}}_2}{\overset{\mathrm{\Delta }}{\to }}\text{W} \\ \text{5}{\text{O}}_2+{\text{P}}_4\stackrel{\mathrm{\Delta }}{\to }\text{X} \\ {\text{P}}_4{\text{O}}_{10}\stackrel{\text{HN}{\text{O}}_3}{\to }\text{Y}+\text{Z} \end{gathered}$$
We have to find the compounds W, X, Y, and Z.
When $\text{KCl}{\text{O}}_3$ is heated in the presence of a catalyst, $\text{Mn}{\text{O}}_2$, oxygen gas is formed. The reaction is given below:
$$3\text{KCl}{\text{O}}_3+\text{Mn}{\text{O}}_2\stackrel{\mathrm{\Delta }}{\to }3{\text{O}}_2+3\text{KCl}$$
Thus the compound W is oxygen gas, ${\text{O}}_2$.
Next step is the reaction of an excess amount of W with white phosphorus, i.e. ${\text{P}}_4$.
i.e. $3{\text{O}}_2\stackrel{{\text{P}}_4}{\to }{\text{P}}_4{\text{O}}_{10}$
Now we have obtained the compound X, i.e. X is ${\text{P}}_4{\text{O}}_{10}$.
Now we have to react to the compound X with pure nitric acid, $\text{HN}{\text{O}}_3$ to give Y and Z. The corresponding reaction is given below:
${\text{P}}_4{\text{O}}_{10}\stackrel{\text{HN}{\text{O}}_3}{\to }\text{HP}{\text{O}}_3+{\text{N}}_2{\text{O}}_5$
Thus the compounds Y and Z are obtained. Y is ${\text{H}}_3\text{P}{\text{O}}_3$ and Z is ${\text{N}}_2{\text{O}}_5$.
Altogether, the complete chemical balanced equation can be represented as given below:
$3\text{KCl}{\text{O}}_3\underset{\text{catalyst}}{\overset{\text{Mn}{\text{O}}_2}{\to }}3{\text{O}}_2+3\text{KCl}\stackrel{{\text{P}}_4}{\to }{\text{P}}_4{\text{O}}_{10}\stackrel{\text{HN}{\text{O}}_3}{\to }\text{HP}{\text{O}}_3+{\text{N}}_2{\text{O}}_5$
Thus the compounds W, X, Y and Z are ${\text{O}}_2,{\text{P}}_4{\text{O}}_{10},{\text{N}}_2{\text{O}}_5,\text{HP}{\text{O}}_3$ respectively.
(a) W and X are ${\text{O}}_2,{\text{P}}_4{\text{O}}_{10}$ respectively.
Hence the correct option is A.

(b) Y and Z are ${\text{N}}_2{\text{O}}_5$ and $\text{HP}{\text{O}}_3$ respectively.

Hence the correct option is A.

Additional information:
The amount of oxygen reacted with white phosphorus can differ the products formed. If the amount of oxygen gas is less, ${\text{P}}_4{\text{O}}_6$ is formed. When the amount is excess, the product formed will be ${\text{P}}_4{\text{O}}_{10}$ .

Note: When we come across this type of reaction, it is important to write and balance the chemical equation. Such reactions require energy sources like heat, light or electricity which helps to break the bonds between them. The reaction of heating potassium chlorate is an example of decomposition reaction.