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

Upon heating HClO_(3) in the presence of catalytic amount of MnO_(2),

1.	Upon heating HClO_(3) in the presence of catalytic amount of MnO_(2), a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO_(3) gives Y and Z. Y and Z are, respectively
Answer» `N_(2)O_(5) and HPO_(3)` `N_(2)O_(3) and H_(3)PO_(4)` `N_(2)O_(4) and H_(3)PO_(3)` `N_(2)O_(4) and HPO_(3)` Solution :`{:(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))`