What volume of 0.1M `KMnO_(4)` is needed to oxidize 100 mg of `FeC_(2)O_(4)` in acid solution ?

Options

A. 4.1 mL
B. 8.2 mL
C. 10.2 mL
D. 4.6 mL

Explanation

To solve the problem of determining the volume of 0.1 M KMnO4 needed to oxidize 100 mg of FeC2O4 in an acidic solution, we can follow these steps:

### Step 1: Write the balanced chemical reaction
In acidic medium, the reaction between KMnO4 and FeC2O4 can be represented as:
\[ \text{MnO}_4^- + \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + 2\text{CO}_2 \]

### Step 2: Determine the oxidation states and n-factors
- For KMnO4, Mn changes from +7 to +2, so the change in oxidation state is 5. Therefore, the n-factor of KMnO4 is 5.
- For FeC2O4, Fe changes from +2 to +3 (n-factor = 1), and each carbon in C2O4 changes from +3 to +4 (total change = 2 for 2 carbon atoms). Thus, the total n-factor for FeC2O4 is 1 + 2 = 3.

### Step 3: Calculate the number of moles of FeC2O4
Given mass of FeC2O4 = 100 mg = 0.1 g.
Molar mass of FeC2O4 = 143.91 g/mol.

\[ \text{Number of moles of FeC}_2\text{O}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1 \, \text{g}}{143.91 \, \text{g/mol}} = 6.95 \times 10^{-4} \, \text{mol} \]

### Step 4: Calculate the equivalent moles of KMnO4 needed
Using the relation:
\[ \text{n-factor of FeC}_2\text{O}_4 \times \text{moles of FeC}_2\text{O}_4 = \text{n-factor of KMnO}_4 \times \text{moles of KMnO}_4 \]
\[ 3 \times 6.95 \times 10^{-4} = 5 \times \text{moles of KMnO}_4 \]
\[ \text{moles of KMnO}_4 = \frac{3 \times 6.95 \times 10^{-4}}{5} = 4.17 \times 10^{-4} \, \text{mol} \]

### Step 5: Calculate the volume of KMnO4 solution required
Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} = \frac{4.17 \times 10^{-4}}{0.1} = 4.17 \times 10^{-3} \, \text{L} \]

### Step 6: Convert volume from liters to milliliters
\[ \text{Volume (mL)} = 4.17 \times 10^{-3} \, \text{L} \times 1000 \, \text{mL/L} = 4.17 \, \text{mL} \]

### Final Answer
The volume of 0.1 M KMnO4 needed to oxidize 100 mg of FeC2O4 in acidic solution is approximately **4.17 mL**.

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