What volume of `O_(2)` measured at standard condition will be formed by the action of `100 mL` of `0.5N KMnO_(4)` on hydrogen peroxide in an acid solution?
The skeleton equation for the reaction is,
`KMnO_(4)+H_(2)SO_(4)+H_(2)O_(2)rarrK_2SO_(4)+MnSO_(4)+H_(2)O+O_(2)`

Options

A. 0.12litre
B. 0.028 litre
C. 0.56 litre
D. 1.12 litre

Answer

To solve the problem of finding the volume of \( O_2 \) produced from the reaction of \( 100 \, \text{mL} \) of \( 0.5 \, N \, KMnO_4 \) with hydrogen peroxide in an acid solution, we will follow these steps:

### Step 1: Calculate the milliequivalents of \( KMnO_4 \)

The formula for calculating milliequivalents is:

\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (in L)} \]

Given:
- Normality of \( KMnO_4 = 0.5 \, N \)
- Volume of \( KMnO_4 = 100 \, \text{mL} = 0.1 \, \text{L} \)

Calculating milliequivalents:

\[ \text{Milliequivalents of } KMnO_4 = 0.5 \, N \times 0.1 \, \text{L} = 0.05 \, \text{equivalents} \]

### Step 2: Equate the milliequivalents of \( KMnO_4 \) to \( O_2 \)

From the reaction, we can see that the milliequivalents of \( KMnO_4 \) will equal the milliequivalents of \( O_2 \) produced.

\[ \text{Milliequivalents of } O_2 = 0.05 \, \text{equivalents} \]

### Step 3: Determine the equivalent weight of \( O_2 \)

The equivalent weight is calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]

For \( O_2 \):
- Molar mass of \( O_2 = 32 \, g/mol \)
- The change in oxidation state for \( O_2 \) is from \(-1\) (in \( H_2O_2 \)) to \( 0 \), which gives \( n = 2 \).

Calculating equivalent weight:

\[ \text{Equivalent Weight of } O_2 = \frac{32 \, g/mol}{2} = 16 \, g/equiv \]

### Step 4: Calculate the weight of \( O_2 \)

Using the milliequivalents calculated earlier:

\[ \text{Weight of } O_2 = \text{Milliequivalents} \times \text{Equivalent Weight} \]

\[ \text{Weight of } O_2 = 0.05 \, \text{equivalents} \times 16 \, g/equiv = 0.8 \, g \]

### Step 5: Calculate the volume of \( O_2 \) produced at STP

At standard temperature and pressure (STP), \( 1 \, \text{mol} \) of gas occupies \( 22.4 \, L \).

Using the formula:

\[ \text{Volume of } O_2 = \frac{\text{Weight of } O_2}{\text{Molar Mass of } O_2} \times 22.4 \, L \]

Calculating volume:

\[ \text{Volume of } O_2 = \frac{0.8 \, g}{32 \, g/mol} \times 22.4 \, L \]

\[ \text{Volume of } O_2 = 0.025 \, mol \times 22.4 \, L/mol = 0.56 \, L \]

### Final Answer

The volume of \( O_2 \) produced at standard conditions is \( 0.56 \, L \).

---