Upon heating`KClO_(3)` in presence of catalytic amount of `MnO_(2)`, a gas `W` is formed. Excess amount of `W` reacts with white phosphorus to given `X`. The reaction of `X` with pure `HNO_(3)` gives `Y` and `Z`.
Y and Z are, respectively

Options

A. `N_(2)O_(5) and HPO_(3)`
B. `N_(2)O_(3) and H_(3)PO_(4)`
C. `N_(2)O_(4) and H_(3)PO_(3)`
D. `N_(2)O_(4) and HPO_(3)`

Answer

A

Explanation

`{:(2KClo_(3) overset(MnO_(2)"," Delta)rarr 2 KCl + underset((W))(3O_(2))),(P_(4)+10O_(2) overset("Oxidation")rarr underset((X))(P_(4)O_(10))):}`
`P_(4)O_(10)` acts as a dehydrating agent and removes `H_(2)O` from `HNO_(3)` to give `N_(2)O_(5)` (Y) and itself gets converted into `HPO_(3)` (Z).
Thus, X is `P_(4)O_(10)`, Y is `N_(2)O_(5)`, and Z in `HPO_(3)` and W is `O_(2)`.
`P_(4)O_(10)+4HNO_(3)rarr underset((Y))(2N_(2)O_(5))+underset((Z))(4HPO_(3))`