During thhe preparation of `H_(2)S_(2)O_(8)` (per disulphuric acid) `O_(2)` gas also releases at anode as byproduct. When 9.72 of `H_(2)` releases at cathode and `2.35`L `O_(2)` at anode at STP, the weight of `H_(2)S_(2)O_(8)` produced in gram is

Options

A. 87.12
B. 43.56
C. 83.42
D. 51.74

Explanation

To solve the problem step by step, we will use the information given about the volumes of hydrogen and oxygen released at STP and the stoichiometry of the reaction involved in the preparation of per disulfuric acid (H₂S₂O₈).

### Step 1: Understand the Reaction
The reaction for the preparation of per disulfuric acid is:
\[ \text{H}_2\text{S}_2\text{O}_8 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 + \text{H}_2 + \text{O}_2 \]

### Step 2: Calculate Moles of Gases
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. We need to calculate the moles of hydrogen (H₂) and oxygen (O₂) produced.

- For hydrogen (H₂):
\[ \text{Moles of H}_2 = \frac{9.72 \text{ L}}{22.4 \text{ L/mol}} = 0.433 \text{ moles} \]

- For oxygen (O₂):
\[ \text{Moles of O}_2 = \frac{2.35 \text{ L}}{22.4 \text{ L/mol}} = 0.105 \text{ moles} \]

### Step 3: Determine the Stoichiometry
From the balanced reaction, we see that:
- 1 mole of H₂S₂O₈ produces 1 mole of H₂ and 1 mole of O₂.
- Therefore, the moles of H₂ and O₂ produced can be related to the moles of H₂S₂O₈.

Let \( x \) be the moles of H₂S₂O₈ produced. Then:
\[ x = \text{moles of H}_2 = 0.433 \text{ moles} \]
\[ x = \text{moles of O}_2 = 0.105 \text{ moles} \]

### Step 4: Use the Stoichiometric Relationships
From the reaction, we can find the relationship between the moles of H₂S₂O₈ produced and the moles of H₂ and O₂:
\[ \text{Moles of H}_2 = \text{Moles of H}_2S_2O_8 \]
\[ \text{Moles of O}_2 = \frac{1}{2} \times \text{Moles of H}_2S_2O_8 \]

### Step 5: Calculate the Moles of H₂S₂O₈
Using the moles of O₂ produced:
\[ 0.105 = \frac{1}{2} x \implies x = 0.21 \text{ moles of H}_2S_2O_8 \]

### Step 6: Calculate the Mass of H₂S₂O₈
The molar mass of H₂S₂O₈ is 194 g/mol. Therefore, the mass produced is:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.21 \text{ moles} \times 194 \text{ g/mol} = 40.94 \text{ g} \]

### Step 7: Round the Answer
Rounding to two decimal places, the weight of H₂S₂O₈ produced is approximately:
\[ \text{Weight} \approx 40.94 \text{ g} \]

### Final Answer
The weight of H₂S₂O₈ produced is approximately **40.94 grams**.

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