During electrolysis of `H_2SO_4`(aq) with high charge density, `H_2 S_2O_8` is fromed as by product In such electrolysis 44.8 `LH_2(g)` and 15 `L O_2(g)` liberated at STP Hence, the moles of `H_2 S_2 O_8` formed is approximately equal to

Options

A. 0.25
B. 0.66
C. 2
D. 2.68

Answer

To solve the problem, we need to calculate the moles of `H2S2O8` formed during the electrolysis of `H2SO4` based on the volumes of hydrogen and oxygen gas liberated.

### Step-by-Step Solution:

1. **Calculate Moles of Hydrogen Gas (H2)**:
- Given volume of `H2` = 44.8 L
- At STP, 1 mole of gas occupies 22.4 L.
- Therefore, moles of `H2` = Volume of `H2` / Molar Volume = 44.8 L / 22.4 L/mol = 2 moles.

2. **Calculate Moles of Oxygen Gas (O2)**:
- Given volume of `O2` = 15 L
- Using the same molar volume at STP:
- Moles of `O2` = Volume of `O2` / Molar Volume = 15 L / 22.4 L/mol ≈ 0.669 moles.

3. **Determine the Relationship Between Moles of Gases and Moles of H2S2O8**:
- From the electrolysis of `H2SO4`, we know that:
- 2 moles of `H2SO4` produce 2 moles of `H2` and 1 mole of `O2` and 1 mole of `H2S2O8`.
- Thus, the stoichiometry can be represented as:
- 2 moles of `H2` + 1 mole of `O2` + 1 mole of `H2S2O8`.

4. **Setting Up the Equation**:
- Let `n` be the moles of `H2S2O8` formed.
- From the stoichiometry, we can derive:
- Moles of `H2` = 2 * n
- Moles of `O2` = n
- Therefore, we have the following equations:
- 2n = 2 (from `H2`)
- n = 0.669 (from `O2`)

5. **Solving for n**:
- From the `H2` equation, we find n = 1.
- From the `O2` equation, we find n = 0.669.
- Since `H2S2O8` is produced in a 1:1 ratio with `O2`, we take the moles of `O2` to find the moles of `H2S2O8` formed.
- Thus, `n` (moles of `H2S2O8`) ≈ 0.669.

### Final Answer:
The moles of `H2S2O8` formed is approximately equal to **0.669 moles**.