To solve the problem, we first need to understand the stoichiometry of the reaction and how to calculate the mass of peroxy-disulphuric acid (\( \text{H}_2\text{S}_2\text{O}_8 \)) formed during the electrolysis.
### Step 1: Calculate the moles of \( \text{H}_2 \) and \( \text{O}_2 \)
At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of hydrogen and oxygen generated:
- **For \( \text{H}_2 \)**:
\[
\text{Moles of } \text{H}_2 = \frac{\text{Volume of } \text{H}_2}{\text{Molar Volume}} = \frac{9.72 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.433 \, \text{moles}
\]
- **For \( \text{O}_2 \)**:
\[
\text{Moles of } \text{O}_2 = \frac{\text{Volume of } \text{O}_2}{\text{Molar Volume}} = \frac{2.35 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.105 \, \text{moles}
\]
### Step 2: Use the stoichiometry of the reaction
The reaction for the electrolysis is:
\[
2 \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_8 + 2 \text{H}^+ + 2 \text{e}^-
\]
From the equation, we can see that for every 2 moles of \( \text{H}_2\text{SO}_4 \), 1 mole of \( \text{H}_2\text{S}_2\text{O}_8 \) is produced.
### Step 3: Determine the limiting reactant
The formation of \( \text{H}_2 \) and \( \text{O}_2 \) is associated with the reaction. The half-reactions are:
- For \( \text{H}_2 \) generation: \( 2 \text{H}^+ + 2 \text{e}^- \rightarrow \text{H}_2 \)
- For \( \text{O}_2 \) generation: \( 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4 \text{e}^- \)
From the half-reactions:
- 2 moles of \( \text{H}_2 \) require 2 moles of electrons.
- 1 mole of \( \text{O}_2 \) requires 4 moles of electrons.
This means:
- From 0.433 moles of \( \text{H}_2 \), the moles of electrons consumed is \( 0.433 \, \text{moles} \times 2 = 0.866 \, \text{moles of electrons} \).
- From 0.105 moles of \( \text{O}_2 \), the moles of electrons consumed is \( 0.105 \, \text{moles} \times 4 = 0.420 \, \text{moles of electrons} \).
Since \( \text{O}_2 \) requires fewer moles of electrons, it is the limiting reactant.
### Step 4: Calculate moles of \( \text{H}_2\text{S}_2\text{O}_8 \) produced
From the reaction stoichiometry:
- 1 mole of \( \text{H}_2\text{S}_2\text{O}_8 \) is produced for every 4 moles of electrons used (for \( \text{O}_2 \)).
Since 0.105 moles of \( \text{O}_2 \) corresponds to 0.420 moles of electrons, we find:
\[
\text{Moles of } \text{H}_2\text{S}_2\text{O}_8 = \frac{0.420 \, \text{moles of electrons}}{4} \approx 0.105 \, \text{moles}
\]
### Step 5: Calculate the mass of \( \text{H}_2\text{S}_2\text{O}_8 \)
The molar mass of \( \text{H}_2\text{S}_2\text{O}_8 \) is calculated as follows:
- \( 2 \times \text{H} = 2 \times 1.01 = 2.02 \, \text{g/mol} \)
- \( 2 \times \text{S} = 2 \times 32.07 = 64.14 \, \text{g/mol} \)
- \( 8 \times \text{O} = 8 \times 16.00 = 128.00 \, \text{g/mol} \)
Total molar mass:
\[
\text{Molar mass} = 2.02 + 64.14 + 128.00 = 194.16 \, \text{g/mol}
\]
Now, calculate the mass of \( \text{H}_2\text{S}_2\text{O}_8 \) produced:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.105 \, \text{moles} \times 194.16 \, \text{g/mol} \approx 20.40 \, \text{g}
\]
### Final Answer
The mass of peroxy-disulphuric acid formed is approximately **20.40 grams**.
