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

Answer

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

### Step 1: Heating KClO₃ in the presence of MnO₂
When potassium chlorate (KClO₃) is heated in the presence of a catalyst like manganese dioxide (MnO₂), it decomposes to form potassium chloride (KCl) and oxygen gas (O₂).

**Chemical Reaction:**
\[ 2 \, KClO_3 \xrightarrow{MnO_2} 2 \, KCl + 3 \, O_2 \]

**Gas W:**
From this reaction, we can identify that the gas W formed is **O₂** (oxygen gas).

### Step 2: Reaction of W (O₂) with white phosphorus (P₄)
Next, we take the excess amount of oxygen gas (W) and react it with white phosphorus (P₄). The reaction of white phosphorus with oxygen leads to the formation of phosphorus pentoxide (P₄O₁₀).

**Chemical Reaction:**
\[ P_4 + 5 \, O_2 \rightarrow P_4O_{10} \]

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

### Step 3: Reaction of X (P₄O₁₀) with pure HNO₃
Now, we take the product X (P₄O₁₀) and react it with pure nitric acid (HNO₃). The reaction produces dinitrogen pentoxide (N₂O₅) and phosphoric acid (H₃PO₄).

**Chemical Reaction:**
\[ P_4O_{10} + 4 \, HNO_3 \rightarrow 2 \, N_2O_5 + 2 \, H_3PO_4 \]

**Products Y and Z:**
From this reaction, we can identify that:
- Y is **N₂O₅** (dinitrogen pentoxide)
- Z is **H₃PO₄** (phosphoric acid)

### Final Summary
In summary, we have:
- W = O₂ (oxygen gas)
- X = P₄O₁₀ (phosphorus pentoxide)
- Y = N₂O₅ (dinitrogen pentoxide)
- Z = H₃PO₄ (phosphoric acid)

### Answer
Thus, the correct answer for the question is:
- Y and Z are respectively **N₂O₅ and H₃PO₄**.