In the following reaction, `MnO_(2) + 4HCl rarr MnCl_(2) + 2H_(2)O + Cl_(2)`
2 mole `MnO_(2)` reacts with 4 mol of HCl to form 11.2 L `Cl_(2)` at STP. Thus, percentage yield of `Cl_(2)` is:

Options

A. 25%
B. `50%`
C. `100%`
D. 75%

Explanation

To solve the problem, we need to determine the percentage yield of Cl₂ produced in the reaction:

**Given Reaction:**
\[ \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \]

### Step 1: Determine the moles of reactants
From the question, we know:
- 2 moles of MnO₂
- 4 moles of HCl

### Step 2: Identify the limiting reagent
To find the limiting reagent, we compare the mole ratio of the reactants to the coefficients in the balanced equation.

- For MnO₂:
\[ \text{Mole ratio} = \frac{2 \text{ moles}}{1} = 2 \]

- For HCl:
\[ \text{Mole ratio} = \frac{4 \text{ moles}}{4} = 1 \]

Since the ratio for HCl is smaller, HCl is the limiting reagent.

### Step 3: Calculate the theoretical yield of Cl₂
According to the balanced equation, 4 moles of HCl produce 1 mole of Cl₂.

Thus, from 4 moles of HCl, the theoretical yield of Cl₂ is:
\[ \text{Moles of Cl₂ produced} = \frac{4 \text{ moles HCl}}{4} = 1 \text{ mole Cl₂} \]

### Step 4: Convert the theoretical yield of Cl₂ to volume at STP
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the theoretical volume of Cl₂ is:
\[ \text{Volume of Cl₂} = 1 \text{ mole} \times 22.4 \text{ L/mole} = 22.4 \text{ L} \]

### Step 5: Compare with the actual yield
The actual yield of Cl₂ given in the question is 11.2 L.

### Step 6: Calculate the percentage yield
Percentage yield is calculated using the formula:
\[ \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage Yield} = \left( \frac{11.2 \text{ L}}{22.4 \text{ L}} \right) \times 100 = 50\% \]

### Final Answer:
The percentage yield of Cl₂ is **50%**.

---