In the following reaction, `MnO_2 +4HCl rarr MnCl_2 +2H_2O+CI_2` 2mole of `MnO_2` react with 4 moles of HCl to form 11.2L `Cl_2 ` at STP.
Thus, percent yield of `Cl_2`is

Options

A. 0.25
B. 0.5
C. 1
D. 0.75

Answer

To find the percent yield of \( Cl_2 \) from the reaction \( MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2 \), we will follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced equation for the reaction is:
\[ MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2 \]

### Step 2: Determine the moles of \( Cl_2 \) produced theoretically
From the balanced equation, we see that 1 mole of \( MnO_2 \) produces 1 mole of \( Cl_2 \). Therefore, 2 moles of \( MnO_2 \) will produce 2 moles of \( Cl_2 \).

### Step 3: Calculate the volume of \( Cl_2 \) produced at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, 2 moles of \( Cl_2 \) would occupy:
\[ 2 \text{ moles} \times 22.4 \text{ L/mole} = 44.8 \text{ L} \]

### Step 4: Identify the actual yield of \( Cl_2 \)
According to the problem, the actual yield of \( Cl_2 \) produced is 11.2 L.

### Step 5: Calculate the percent yield
The percent yield can be calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
Substituting the values:
\[ \text{Percent Yield} = \left( \frac{11.2 \text{ L}}{44.8 \text{ L}} \right) \times 100 \]
\[ \text{Percent Yield} = \left( \frac{11.2}{44.8} \right) \times 100 = 25\% \]

### Final Answer
The percent yield of \( Cl_2 \) is **25%**.

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