During the electrolysis of conc `H_(2)SO_(4)`, it was found that `H_(2)S_(2)O_(8)` and `O_(2)` liberated in a molar ratio of `3:1`. How many moles of `H_(2)` were found of moles of `H_(2)S_(2)O_(8)` ?
`(` Express your answer as `:3xx mol e s of H_(2),` integer answer is between 0 and 5`0`

Explanation

To solve the problem, we need to determine how many moles of hydrogen gas (H₂) are produced in relation to the moles of disulfuric acid (H₂S₂O₈) formed during the electrolysis of concentrated sulfuric acid (H₂SO₄). The problem states that H₂S₂O₈ and O₂ are liberated in a molar ratio of 3:1.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the number of moles of \( O_2 \) produced. According to the problem, the moles of \( H_2S_2O_8 \) produced will be \( 3x \) since they are in a 3:1 ratio.

2. **Identify Reactions**:
- **Cathode Reaction**: At the cathode, water is reduced to produce hydrogen gas:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
- **Anode Reactions**: At the anode, two reactions can occur:
1. Oxidation of water to oxygen gas:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
2. Oxidation of sulfate ions to form \( H_2S_2O_8 \):
\[ 2SO_4^{2-} \rightarrow H_2S_2O_8 + 2e^- \]

3. **Calculate Faraday Requirements**:
- For \( x \) moles of \( O_2 \) produced, the total charge required is \( 4x \) Faraday (since 1 mole of \( O_2 \) requires 4 Faraday).
- For \( 3x \) moles of \( H_2S_2O_8 \) produced, the total charge required is \( 6x \) Faraday (since 1 mole of \( H_2S_2O_8 \) requires 2 Faraday).

4. **Total Faraday Calculation**:
The total charge (Faraday) used at the anode is:
\[ \text{Total Faraday} = 4x + 6x = 10x \]

5. **Relate Faraday to Hydrogen Production**:
At the cathode, 1 mole of hydrogen gas is produced by 2 Faraday. Therefore, the moles of hydrogen produced can be calculated as:
\[ \text{Moles of } H_2 = \frac{10x}{2} = 5x \]

6. **Find Moles of Hydrogen per Moles of \( H_2S_2O_8 \)**:
We need to express the moles of hydrogen in terms of the moles of \( H_2S_2O_8 \):
\[ \text{Moles of } H_2 \text{ per moles of } H_2S_2O_8 = \frac{5x}{3x} = \frac{5}{3} \]

7. **Final Answer Format**:
According to the question, we need to express the answer in the form of \( 3xx \) moles of \( H_2 \). Therefore, we multiply \( \frac{5}{3} \) by 3:
\[ 3 \times \frac{5}{3} = 5 \]

Thus, the final answer is:
\[ \text{5 moles of } H_2 \]