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

1.	Upon heating KClO_(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)andHPO_(3)` `N_(2)O_(3)andH_(3)PO_(4)` `N_(2)O_(4)andH_(3)PO_(3)` `N_(2)O_(4)andHPO_(3)` Solution :`2KClO_(3)overset(MnO_(2),Delta)to2KCl+underset((W))(3O_(3))` `P_(4)+10O_(2)overset("Oxidation")tounderset((X))(P_(4)O_(10))` `P_(4)O_(10)` acts as 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)tounderset((Y))(2N_(2)O_(5))+underset((Z))(4HPO_(3))`