How many mL of 22.4 volume `H_(2)O_(2)` is required to oxidise 0.1 mol of `H_(2)S` gas to S ?

Answer

To solve the problem of how many mL of 22.4 volume H₂O₂ is required to oxidize 0.1 mol of H₂S gas to sulfur, we can break it down into several steps.

### Step 1: Understand the Reaction
The oxidation of hydrogen sulfide (H₂S) to sulfur (S) can be represented by the following reaction:
\[ \text{H}_2\text{S} + \text{H}_2\text{O}_2 \rightarrow \text{S} + \text{H}_2\text{O} \]

### Step 2: Determine the Stoichiometry
From the balanced reaction, we can see that 1 mole of H₂S reacts with 1 mole of H₂O₂. Therefore, if we have 0.1 moles of H₂S, we will need 0.1 moles of H₂O₂.

### Step 3: Calculate the Normality of H₂O₂
The volume strength of H₂O₂ is given as 22.4. The relationship between volume strength (VS) and normality (N) is:
\[ \text{VS} = \text{N} \times 5.6 \]
Using this, we can calculate the normality of the H₂O₂ solution:
\[ \text{N} = \frac{\text{VS}}{5.6} = \frac{22.4}{5.6} = 4 \, \text{N} \]

### Step 4: Calculate the Number of Equivalents
The number of equivalents of H₂S can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Number of moles} \times \text{n-factor} \]
For H₂S, the n-factor is 2 (since it loses 2 hydrogen atoms to form sulfur). Therefore:
\[ \text{Number of equivalents of H}_2\text{S} = 0.1 \, \text{mol} \times 2 = 0.2 \, \text{equivalents} \]

### Step 5: Calculate the Volume of H₂O₂ Required
Using the equivalence relationship:
\[ \text{Volume of H}_2\text{O}_2 \, (\text{in L}) = \frac{\text{Number of equivalents}}{\text{Normality}} \]
Substituting the known values:
\[ \text{Volume of H}_2\text{O}_2 = \frac{0.2 \, \text{equivalents}}{4 \, \text{N}} = 0.05 \, \text{L} \]

### Step 6: Convert to mL
To convert liters to milliliters:
\[ 0.05 \, \text{L} = 0.05 \times 1000 \, \text{mL} = 50 \, \text{mL} \]

### Final Answer
Therefore, the volume of 22.4 volume H₂O₂ required to oxidize 0.1 mol of H₂S gas to sulfur is **50 mL**.

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