`KCIO_(3)` on reaction with `SO_(2)` gives (A) and on reactionwith conc. `H_(2)SO_(4)` gives (B).(A) and (B) are

Options

A. `KCI,HCI_(4)`
B. `CI_(2),CIO_(2)`
C. `KCI,HCIO_(3)`
D. `CI_(2),HCIO_(4)`

Explanation

To solve the question, we need to analyze the reactions of potassium chlorate (KClO₃) with sulfur dioxide (SO₂) and concentrated sulfuric acid (H₂SO₄).

### Step 1: Reaction of KClO₃ with SO₂
When KClO₃ reacts with SO₂, the reaction can be represented as follows:

\[ \text{KClO}_3 + \text{SO}_2 \rightarrow \text{KCl} + \text{SO}_3 \]

In this reaction, potassium chlorate (KClO₃) is reduced, and sulfur dioxide (SO₂) acts as a reducing agent. The product formed from this reaction is potassium chloride (KCl).

**Product A:** KCl

### Step 2: Reaction of KClO₃ with concentrated H₂SO₄
Now, let's consider the reaction of KClO₃ with concentrated sulfuric acid (H₂SO₄). The reaction can be represented as follows:

\[ \text{KClO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{HClO}_4 + \text{KHSO}_4 + \text{ClO}_2 + \text{H}_2\text{O} \]

In this reaction, potassium chlorate reacts with concentrated sulfuric acid to produce several products, including perchloric acid (HClO₄).

**Product B:** HClO₄

### Conclusion
From the reactions we analyzed:
- Product A (from the reaction with SO₂) is KCl.
- Product B (from the reaction with concentrated H₂SO₄) is HClO₄.

Thus, the final answer is:
- (A) = KCl
- (B) = HClO₄