What 'X' in the following reaction ?
`KHSO_(4)+F_(2) to HF+X`

Options

A. `K_(2)SO_(4)`
B. `K_(2)S_(2)SO_(4)`
C. `K_(2)S_(2)O_(8)`
D. `K_(2)S_(2)O_(2)`

Explanation

To solve the question regarding the reaction of KHSO4 with F2 to produce HF and an unknown substance 'X', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reactants given are potassium hydrogen sulfate (KHSO4) and fluorine gas (F2). The products include hydrofluoric acid (HF) and an unknown substance 'X'.

2. **Write the Reaction**:
The reaction can be represented as:
\[ KHSO_4 + F_2 \rightarrow HF + X \]

3. **Determine the Product 'X'**:
When KHSO4 reacts with fluorine, it undergoes a transformation. The fluorine is a strong oxidizing agent and can react with the sulfate ion present in KHSO4. The expected product from this reaction is dipotassium peroxodisulfate, which has the formula K2S2O8.

4. **Balance the Reaction**:
To balance the reaction, we can write:
\[ 2 KHSO_4 + F_2 \rightarrow 2 HF + K_2S_2O_8 \]
This shows that two moles of KHSO4 react with one mole of F2 to produce two moles of HF and one mole of K2S2O8.

5. **Identify 'X'**:
From the balanced equation, we can see that 'X' corresponds to K2S2O8.

### Conclusion:
Thus, the substance 'X' in the reaction is K2S2O8 (dipotassium peroxodisulfate).