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

Options

A. `O_(2) and P_(4)O_(10)`
B. `O_(2) and P_(4)O_(6)`
C. `O_(3) and P_(4)O_(6)`
D. `O_(3) and P_(4)O_(10)`

Answer

To solve the question step by step, we will analyze each part of the reaction described.

### Step 1: Identify the gas W formed from the heating of KClO3
When potassium chlorate (KClO3) is heated in the presence of manganese dioxide (MnO2), it decomposes to produce potassium chloride (KCl) and oxygen gas (O2).

**Equation:**
\[ 2KClO_3 \xrightarrow{MnO_2} 2KCl + 3O_2 \]

**Conclusion:**
Thus, the gas W formed is **O2 (oxygen gas)**.

### Step 2: Determine the reaction of W (O2) with white phosphorus
White phosphorus (P4) reacts with excess oxygen (W) to form phosphorus pentoxide (P4O10).

**Equation:**
\[ P_4 + 5O_2 \rightarrow P_4O_{10} \]

**Conclusion:**
Therefore, the product X formed from this reaction is **P4O10 (phosphorus pentoxide)**.

### Step 3: Identify the reaction of X (P4O10) with pure HNO3
When phosphorus pentoxide (P4O10) reacts with nitric acid (HNO3), it produces nitrogen pentoxide (N2O5) and phosphoric acid (H3PO4).

**Equation:**
\[ P_4O_{10} + 6HNO_3 \rightarrow 2N_2O_5 + 4H_3PO_4 \]

**Conclusion:**
From this reaction, we can identify Y as **N2O5** and Z as **H3PO4**.

### Final Summary:
- W (gas formed) = **O2**
- X (product formed from reaction with white phosphorus) = **P4O10**