10g of` MnO_(2)`on reactoin with HCI forms 2.24 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%

Answer

To solve the problem of determining the percentage impurity of MnO₂ in a sample that weighs 10 g, we will follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \]
This equation shows that 1 mole of MnO₂ produces 1 mole of Cl₂.

2. **Calculate the number of moles of Cl₂ produced:**
Given that 2.24 L of Cl₂ is produced at NTP (Normal Temperature and Pressure), we can use the molar volume of a gas at NTP, which is 22.4 L/mol.
\[ \text{Number of moles of Cl}_2 = \frac{\text{Volume of Cl}_2}{\text{Molar volume at NTP}} = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ moles} \]

3. **Determine the mass of MnO₂ that reacted:**
From the balanced equation, we know that 1 mole of MnO₂ produces 1 mole of Cl₂. Therefore, 0.1 moles of Cl₂ would require:
\[ \text{Mass of MnO}_2 = \text{Moles of MnO}_2 \times \text{Molar mass of MnO}_2 \]
The molar mass of MnO₂ is 87 g/mol.
\[ \text{Mass of MnO}_2 = 0.1 \text{ moles} \times 87 \text{ g/mol} = 8.7 \text{ g} \]

4. **Calculate the mass of impurity in the sample:**
The initial mass of the sample is 10 g. The mass of MnO₂ that reacted is 8.7 g, so the mass of impurity is:
\[ \text{Mass of impurity} = \text{Total mass} - \text{Mass of MnO}_2 = 10 \text{ g} - 8.7 \text{ g} = 1.3 \text{ g} \]

5. **Calculate the percentage impurity:**
The percentage impurity can be calculated using the formula:
\[ \text{Percentage impurity} = \left( \frac{\text{Mass of impurity}}{\text{Total mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage impurity} = \left( \frac{1.3 \text{ g}}{10 \text{ g}} \right) \times 100 = 13\% \]

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