If `100 mL` of acidified `2NH_(2)O_(2)` is allowed to react with `KMnO_(4)` solution till there is light tinge of purples colour, the volume of oxygen produced at `STP` is :

Options

A. 2.24 L
B. 1.12 L
C. 3.36 L
D. 4.48 L

Explanation

To solve the problem of determining the volume of oxygen produced at STP when 100 mL of acidified 2N H₂O₂ reacts with KMnO₄, we can follow these steps:

### Step 1: Understand the Reaction
In this reaction, hydrogen peroxide (H₂O₂) acts as a reducing agent and reacts with potassium permanganate (KMnO₄), which acts as an oxidizing agent. During this reaction, oxygen (O₂) is produced.

### Step 2: Calculate the Volume of O₂ from H₂O₂
We know that 1N H₂O₂ produces 5.6 liters of O₂ at STP. Since we have a 2N solution, the volume of O₂ produced will be double that of a 1N solution.

\[ \text{Volume of O₂ from 1N H₂O₂} = 5.6 \, \text{L} \]
\[ \text{Volume of O₂ from 2N H₂O₂} = 2 \times 5.6 \, \text{L} = 11.2 \, \text{L} \]

### Step 3: Relate the Volume of H₂O₂ to the Volume of O₂
We have 100 mL of 2N H₂O₂. To find out how much O₂ is produced from this volume, we can set up a proportion based on the normality and the volume:

\[ \text{Volume of O₂ produced} = \left( \frac{11.2 \, \text{L}}{1000 \, \text{mL}} \right) \times 100 \, \text{mL} = 1.12 \, \text{L} \]

### Step 4: Consider the Reaction with KMnO₄
Since the stoichiometry of the reaction indicates that the amount of O₂ produced by H₂O₂ is equal to the amount produced by KMnO₄, we can conclude that KMnO₄ will also produce the same volume of O₂.

### Step 5: Total Volume of O₂ Produced
The total volume of O₂ produced in the reaction will be the sum of the volumes produced by both H₂O₂ and KMnO₄:

\[ \text{Total Volume of O₂} = 1.12 \, \text{L} + 1.12 \, \text{L} = 2.24 \, \text{L} \]

### Final Answer
Thus, the total volume of oxygen produced at STP is **2.24 liters**.

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