A 0.13 g of a specimen containing `MnO_(2)` is treated with iodide ions. If iodine liberated requires 30.0 " mL of " 0.075 M solution of `Na_(2)S_(2)O_(3)`, the percentage of `MnO_(2)` in the mineral is

Options

A. `75.3%`
B. `85.3%`
C. `95.3%`
D. none

Explanation

To solve the problem step by step, we will follow the chemical reactions involved and the stoichiometric calculations.

### Step 1: Write the reactions involved
The reactions involved in the process are:
1. \( \text{MnO}_2 + 2 \text{I}^- \rightarrow \text{I}_2 + \text{Mn}^{2+} \)
2. \( \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \)

### Step 2: Calculate the milliequivalents of \( \text{Na}_2\text{S}_2\text{O}_3 \)
The formula for calculating milliequivalents is:
\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n factor} \]
Given:
- Molarity of \( \text{Na}_2\text{S}_2\text{O}_3 = 0.075 \, \text{M} \)
- Volume = \( 30.0 \, \text{mL} = 0.030 \, \text{L} \)
- n factor for \( \text{Na}_2\text{S}_2\text{O}_3 = 1 \)

Calculating milliequivalents:
\[ \text{Milliequivalents of } \text{Na}_2\text{S}_2\text{O}_3 = 0.075 \times 0.030 \times 1 = 0.00225 \, \text{equivalents} \]
Converting to milliequivalents:
\[ 0.00225 \, \text{equivalents} = 2.25 \, \text{milliequivalents} \]

### Step 3: Relate milliequivalents of \( \text{Na}_2\text{S}_2\text{O}_3 \) to \( \text{I}_2 \) and \( \text{MnO}_2 \)
From the stoichiometry of the reactions:
- 1 mole of \( \text{I}_2 \) reacts with 2 moles of \( \text{Na}_2\text{S}_2\text{O}_3 \)
- Therefore, the milliequivalents of \( \text{I}_2 \) will also be equal to the milliequivalents of \( \text{MnO}_2 \).

Thus, we have:
\[ \text{Milliequivalents of } \text{MnO}_2 = 2.25 \, \text{milliequivalents} \]

### Step 4: Calculate the weight of \( \text{MnO}_2 \)
The equivalent weight of \( \text{MnO}_2 \) can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n factor}} \]
The molecular weight of \( \text{MnO}_2 \) is approximately \( 87 \, \text{g/mol} \) and the n factor is \( 2 \):
\[ \text{Equivalent weight of } \text{MnO}_2 = \frac{87}{2} = 43.5 \, \text{g/equiv} \]

Now, using the milliequivalents calculated:
\[ \text{Weight of } \text{MnO}_2 = \text{Milliequivalents} \times \text{Equivalent weight} = 2.25 \times 43.5 \, \text{mg} = 98.25 \, \text{mg} = 0.09825 \, \text{g} \]

### Step 5: Calculate the percentage of \( \text{MnO}_2 \) in the specimen
The percentage of \( \text{MnO}_2 \) in the specimen can be calculated as:
\[ \text{Percentage of } \text{MnO}_2 = \left( \frac{\text{Weight of } \text{MnO}_2}{\text{Total weight of specimen}} \right) \times 100 \]
Given the total weight of the specimen is \( 0.13 \, \text{g} \):
\[ \text{Percentage of } \text{MnO}_2 = \left( \frac{0.09825}{0.13} \right) \times 100 \approx 75.19\% \]

### Final Answer
The percentage of \( \text{MnO}_2 \) in the mineral is approximately \( 75.2\% \).

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