Marshell's acid is prepared by the electrolytic oxidation of `H_(2)SO_(4)` as
`2H_(2)SO_(4) rarr H_(2)S_(2)O_(8)+2H^(o+)+2e^(-)`
Oxygen and hydrogen are byproducts. In such electrolysis `2.24L` of `H_(2)` and `0.56L` of `O_(2)` were product at `STP`. The weight of `H_(2)S_(2)O_(8)` fromed is

Options

A. `9.7g`
B. `19.4g`
C. `14.55g`
D. `29.1g`

Explanation

To solve the problem, we need to determine the weight of `H2S2O8` formed during the electrolysis of `H2SO4`, given the volumes of hydrogen and oxygen produced.

### Step-by-Step Solution:

1. **Identify the Reaction:**
The reaction for the electrolysis of sulfuric acid is given as:
\[ 2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^- \]
During this process, hydrogen and oxygen are produced as byproducts.

2. **Calculate Moles of Hydrogen and Oxygen:**
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
- For hydrogen (`H2`):
\[ \text{Volume of } H_2 = 2.24 \, \text{L} \]
\[ \text{Moles of } H_2 = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]

- For oxygen (`O2`):
\[ \text{Volume of } O_2 = 0.56 \, \text{L} \]
\[ \text{Moles of } O_2 = \frac{0.56 \, \text{L}}{22.4 \, \text{L/mol}} = 0.025 \, \text{mol} \]

3. **Determine the Stoichiometry:**
From the electrolysis reaction, we can see that:
- 2 moles of `H2` are produced for every 1 mole of `H2S2O8` formed.
- 1 mole of `O2` is produced for every 1 mole of `H2S2O8` formed.

Therefore, the moles of `H2S2O8` produced can be calculated from the moles of `H2` and `O2`:
- From `H2`:
\[ \text{Moles of } H_2S_2O_8 = \frac{0.1 \, \text{mol } H_2}{2} = 0.05 \, \text{mol} \]
- From `O2`:
\[ \text{Moles of } H_2S_2O_8 = 0.025 \, \text{mol} \]

The limiting factor is `O2`, thus:
\[ \text{Moles of } H_2S_2O_8 = 0.025 \, \text{mol} \]

4. **Calculate the Weight of `H2S2O8`:**
The molecular weight of `H2S2O8` is 194 g/mol. Therefore, the weight can be calculated as:
\[ \text{Weight} = \text{Moles} \times \text{Molecular Weight} \]
\[ \text{Weight} = 0.025 \, \text{mol} \times 194 \, \text{g/mol} = 4.85 \, \text{g} \]

5. **Final Calculation:**
Since we calculated the moles based on the limiting reactant, we must ensure that the calculations are consistent. The weight of `H2S2O8` formed is:
\[ \text{Weight} = 0.05 \, \text{mol} \times 194 \, \text{g/mol} = 9.7 \, \text{g} \]

### Final Answer:
The weight of `H2S2O8` formed is **9.7 grams**.

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