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 (S), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen sulfide (H₂S) and hydrogen peroxide (H₂O₂) can be represented as follows:

\[ H_2S + H_2O_2 \rightarrow S + H_2O \]

### Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we can see that 1 mole of H₂S reacts with 1 mole of H₂O₂. Therefore, the mole ratio is 1:1.

### Step 3: Calculate the moles of H₂O₂ required
Since we have 0.1 moles of H₂S, we will need an equal amount of H₂O₂:

\[ \text{Moles of H₂O₂ required} = 0.1 \text{ mol} \]

### Step 4: Convert volume strength to molarity
The volume strength of H₂O₂ is given as 22.4. We can convert this to molarity using the relationship:

\[ \text{Molarity} = \frac{\text{Volume strength}}{11.2} \]

Substituting the given volume strength:

\[ \text{Molarity of H₂O₂} = \frac{22.4}{11.2} = 2 \text{ M} \]

### Step 5: Calculate the normality of H₂O₂
Since H₂O₂ acts as an oxidizing agent and can provide 2 moles of electrons per mole of H₂O₂, the normality (N) is twice the molarity:

\[ \text{Normality of H₂O₂} = 2 \times 2 = 4 \text{ N} \]

### Step 6: Calculate the milliequivalents of H₂S
The milliequivalents of H₂S can be calculated as:

\[ \text{Milliequivalents of H₂S} = \text{moles} \times \text{valence} \]

The valence of H₂S is 2 (as it can donate 2 electrons):

\[ \text{Milliequivalents of H₂S} = 0.1 \text{ mol} \times 2 = 0.2 \text{ equivalents} = 200 \text{ milliequivalents} \]

### Step 7: Set up the equation for milliequivalents
Since the milliequivalents of H₂O₂ must equal the milliequivalents of H₂S:

\[ \text{Milliequivalents of H₂O₂} = \text{Normality} \times \text{Volume in L} \]

### Step 8: Calculate the volume of H₂O₂ required
We know that:

\[ 200 \text{ meq} = 4 \text{ N} \times V \]

Rearranging gives:

\[ V = \frac{200}{4} = 50 \text{ mL} \]

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

---