10g of` MnO_(2)`on reactoin with HCI forms 2024 L of `Cl_(2)(g) at NTP_(1)` the percentage impurity of `MnO_(2) is :-
MnO_(2)+4HCl to MnCl_(2)+Cl_(2)+2H_(2)O`

Options

A. `87%`
B. `25%`
C. `33.03%`
D. `13%`

Explanation

To solve the problem of determining the percentage impurity of MnO₂ in the given reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The reaction given is:
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \]
From this equation, we can see that 1 mole of MnO₂ produces 1 mole of Cl₂.

2. **Calculate the moles of Cl₂ produced:**
We know that at NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Given that 2.24 liters of Cl₂ is produced:
\[ \text{Moles of Cl}_2 = \frac{\text{Volume of Cl}_2}{22.4 \text{ L}} = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ moles} \]

3. **Determine the moles of MnO₂ required:**
Since the stoichiometry of the reaction shows that 1 mole of MnO₂ produces 1 mole of Cl₂, the moles of MnO₂ required will also be:
\[ \text{Moles of MnO}_2 = 0.1 \text{ moles} \]

4. **Calculate the mass of MnO₂ that corresponds to the moles calculated:**
The molar mass of MnO₂ can be calculated as follows:
- Manganese (Mn) = 55 g/mol
- Oxygen (O) = 16 g/mol
\[ \text{Molar mass of MnO}_2 = 55 + (16 \times 2) = 55 + 32 = 87 \text{ g/mol} \]
Now, we can find the mass of MnO₂ that corresponds to 0.1 moles:
\[ \text{Mass of MnO}_2 = \text{Moles} \times \text{Molar mass} = 0.1 \text{ moles} \times 87 \text{ g/mol} = 8.7 \text{ g} \]

5. **Determine the actual mass of MnO₂ used:**
The problem states that 10 grams of MnO₂ was used in the reaction.

6. **Calculate the percentage purity of MnO₂:**
The percentage purity can be calculated using the formula:
\[ \text{Percentage purity} = \left( \frac{\text{Actual mass of pure MnO}_2}{\text{Total mass of sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage purity} = \left( \frac{8.7 \text{ g}}{10 \text{ g}} \right) \times 100 = 87\% \]

7. **Calculate the percentage impurity:**
The percentage impurity can be calculated as:
\[ \text{Percentage impurity} = 100\% - \text{Percentage purity} = 100\% - 87\% = 13\% \]

### Final Answer:
The percentage impurity of MnO₂ is **13%**.

---