Identify gas $W$ formed from heating $KClO_3$ with $MnO_2$

The well-known decomposition is: $2KClO_3 \xrightarrow[\text{cat. } MnO_2]{\Delta} 2KCl + 3O_2$

So, the gas formed is: $W = O_2$

Reaction of excess $O_2$ with white phosphorus

White phosphorus is $P_4$. Its combustion depends on the amount of oxygen:

• In limited oxygen: $P_4 + 3O_2 \to P_4O_6$
• In excess oxygen: $P_4 + 5O_2 \to P_4O_{10}$

Since the question says excess amount of $W$, the product is: $X = P_4O_{10}$

Check consistency with reaction of $X$ with pure $HNO_3$

$P_4O_{10}$ reacts with concentrated/pure nitric acid to form phosphoryl nitrate and phosphoric acid-related products; this is consistent with $P_4O_{10}$ being the higher oxide formed in excess oxygen.

Thus the identification of $X$ as $P_4O_{10}$ is correct.

Evaluate options

• A: $O_3$ and $P_4O_6$ $\rightarrow$ wrong, since $KClO_3$ gives $O_2$, not $O_3$
• B: $O_2$ and $P_4O_6$ $\rightarrow$ wrong, because oxygen is in excess, so oxide should be $P_4O_{10}$
• C: $O_2$ and $P_4O_{10}$ $\rightarrow$ correct
• D: $O_3$ and $P_4O_{10}$ $\rightarrow$ wrong, since $W \neq O_3$

Final answer

$W = O_2, \quad X = P_4O_{10}$

Hence, the correct option is C.