One of the methods of preparation of per disulphuric acid, `H_(2)S_(2)O_(8)`, involve electrolytic oxidation of `H_(2)SO_(4)` at anode `(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-))` with oxygen ad hydrogen as by-products in such an electrolysis, 9.722 L of `H_(2)` and `2.35L` of `O_(2)` were generated at STP. What is the weighht of `H_(2)S_(2)O_(8)` formed?

Explanation

To solve the problem of determining the weight of perdisulfuric acid (H₂S₂O₈) formed during the electrolysis of H₂SO₄, we can follow these steps:

### Step 1: Understand the Reaction
The reaction at the anode is given as:
\[ 2 \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_8 + 2 \text{H}^+ + 2 e^- \]
This indicates that for every 2 moles of H₂SO₄, 1 mole of H₂S₂O₈ is produced.

### Step 2: Calculate Moles of Hydrogen and Oxygen
At STP, 1 mole of any gas occupies 22.4 L. We can calculate the moles of H₂ and O₂ produced from their volumes.

For H₂:
\[ \text{Volume of H}_2 = 9.722 \, \text{L} \]
\[ \text{Moles of H}_2 = \frac{9.722 \, \text{L}}{22.4 \, \text{L/mol}} = 0.433 \, \text{mol} \]

For O₂:
\[ \text{Volume of O}_2 = 2.35 \, \text{L} \]
\[ \text{Moles of O}_2 = \frac{2.35 \, \text{L}}{22.4 \, \text{L/mol}} = 0.105 \, \text{mol} \]

### Step 3: Determine the Stoichiometry
From the electrolysis reactions:
- 2 moles of H₂O produce 1 mole of O₂ (4 electrons).
- 2 moles of H₂SO₄ produce 1 mole of H₂S₂O₈ (2 electrons).

Using the stoichiometric relationships:
- The moles of H₂ produced correspond to 2 moles of electrons.
- The moles of O₂ produced correspond to 4 moles of electrons.

### Step 4: Set Up the Equation
Using the relationship between the moles of H₂, O₂, and H₂S₂O₈:
\[ \text{Equivalent of H}_2 + \text{Equivalent of O}_2 = \text{Equivalent of H}_2\text{S}_2\text{O}_8 \]

Let \( x \) be the moles of H₂S₂O₈ formed:
\[ 2 \times 0.433 + 4 \times 0.105 = 2x \]
\[ 0.866 + 0.42 = 2x \]
\[ 1.286 = 2x \]
\[ x = \frac{1.286}{2} = 0.643 \, \text{mol} \]

### Step 5: Calculate the Weight of H₂S₂O₈
The molecular weight of H₂S₂O₈ can be calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol × 2 = 64 g/mol
- O: 16 g/mol × 8 = 128 g/mol

Total molecular weight of H₂S₂O₈:
\[ 2 + 64 + 128 = 194 \, \text{g/mol} \]

Now, using the moles calculated:
\[ \text{Weight of H}_2\text{S}_2\text{O}_8 = \text{moles} \times \text{molecular weight} \]
\[ = 0.643 \, \text{mol} \times 194 \, \text{g/mol} = 124.482 \, \text{g} \]

### Final Answer
The weight of H₂S₂O₈ formed is approximately **124.48 g**.

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