Pyrolusite is the main ore of manganese in which it is present as Its Mn content is determined by reducing it under acidic condition to `Mn^(2+)` with the help of oxalate `(C_(2)O_(4)^(2-))` ion which in turn gets oxidized to `CO_(2)` The analytical determination is carried out by adding a known excess volume of `(C_(2)O_(4)^(2-))` solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the `MnO_(2)` has been reduced. The excess unreacted oxalate solution is then titrated with standardized `KMnO_(4)` solution. Thereby Mn content of ore can be calculated. `KMnO_(4)`solution is also standardized under acidic condition against oxalate ion wherein `MnO_(4)^(-)` ion is reduced to `Mn^(2+)` and `(C_(2)O_(4)^(2-))` ion is oxidized to `CO_(2)`
Q of pyrolusite ore were treated with of pure ferrous ammonium sulphate `FeSO_(4).(NH_(4))_(2)SO_(4)6h_(2)O)` and dil. `H_(2)SO_(4)`After the reaction solution was diluted to . of dilute solution required 10 mL of `0.1NK_(2)Cr_(2)O_(7)` solution. What is the amount of ferrous ammonium sulphate in 500 mL of solution, which is neutralized by `K_(2)Cr_(2)O_(7)` solution ?

Options

A. 16.08g
B. 3.92g
C. 19.08g
D. 5.62g

Explanation

To solve the problem, we need to determine the amount of ferrous ammonium sulfate that is neutralized by the potassium dichromate (K₂Cr₂O₇) solution in the given reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The ferrous ammonium sulfate (FeSO₄·(NH₄)₂SO₄·6H₂O) reacts with potassium dichromate (K₂Cr₂O₇) in acidic conditions. The balanced reaction can be represented as:
\[ 6Fe^{2+} + K_2Cr_2O_7 + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O \]
From the equation, we can see that 1 mole of K₂Cr₂O₇ reacts with 6 moles of Fe²⁺.

2. **Calculating Moles of K₂Cr₂O₇**:
Given that 10 mL of 0.1 N K₂Cr₂O₇ is used, we can calculate the number of equivalents of K₂Cr₂O₇:
\[ \text{Equivalents of } K_2Cr_2O_7 = \text{Normality} \times \text{Volume (L)} = 0.1 \, \text{N} \times 0.010 \, \text{L} = 0.001 \, \text{equivalents} \]

3. **Finding Moles of Fe²⁺**:
Since 1 equivalent of K₂Cr₂O₇ reacts with 6 equivalents of Fe²⁺, the equivalents of Fe²⁺ will be:
\[ \text{Equivalents of } Fe^{2+} = 6 \times \text{Equivalents of } K_2Cr_2O_7 = 6 \times 0.001 = 0.006 \, \text{equivalents} \]

4. **Calculating Moles of Fe²⁺**:
The molar mass of Fe²⁺ is approximately 56 g/mol. Therefore, the mass of Fe²⁺ that corresponds to 0.006 equivalents can be calculated as follows:
\[ \text{Moles of } Fe^{2+} = \text{Equivalents} = 0.006 \, \text{moles} \]
\[ \text{Mass of } Fe^{2+} = \text{Moles} \times \text{Molar Mass} = 0.006 \, \text{moles} \times 56 \, \text{g/mol} = 0.336 \, \text{g} \]

5. **Calculating the Amount of Ferrous Ammonium Sulfate**:
The molar mass of ferrous ammonium sulfate (FeSO₄·(NH₄)₂SO₄·6H₂O) is approximately 392 g/mol. Since each mole of ferrous ammonium sulfate contains one mole of Fe²⁺, the mass of ferrous ammonium sulfate required can be calculated as:
\[ \text{Mass of Ferrous Ammonium Sulfate} = \left(\frac{0.336 \, \text{g}}{1 \, \text{mol Fe}^{2+}}\right) \times 392 \, \text{g/mol} = 131.392 \, \text{g} \]

6. **Final Calculation for 500 mL Solution**:
Since the reaction was carried out in a 500 mL solution, the amount of ferrous ammonium sulfate in 500 mL is:
\[ \text{Amount of Ferrous Ammonium Sulfate} = 131.392 \, \text{g} \]

### Conclusion:
The amount of ferrous ammonium sulfate in 500 mL of solution that is neutralized by K₂Cr₂O₇ solution is approximately **131.39 g**.