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 what volume of 0.1 M KMnO4 is needed to oxidize 100 mg of FeC2O4 in an acid solution, we can follow these steps:

### Step 1: Determine the number of moles of FeC2O4
First, we need to calculate the number of moles of FeC2O4 using its mass and molar mass.

- Given mass of FeC2O4 = 100 mg = 0.1 g
- Molar mass of FeC2O4 = 56 (Fe) + 2(12) (C) + 4(16) (O) = 56 + 24 + 64 = 144 g/mol

Now, we can calculate the number of moles:
\[ \text{Number of moles of FeC2O4} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1 \text{ g}}{144 \text{ g/mol}} \approx 0.000694 \text{ mol} \]

### Step 2: Determine the equivalent of FeC2O4
Next, we need to find the number of equivalents of FeC2O4. The n-factor (number of electrons exchanged) for FeC2O4 is 3 (since it can lose 3 electrons).

\[ \text{Equivalents of FeC2O4} = \text{Number of moles} \times \text{n-factor} = 0.000694 \text{ mol} \times 3 \approx 0.00208 \text{ equivalents} \]

### Step 3: Determine the equivalents of KMnO4
In acidic solution, KMnO4 acts as an oxidizing agent and its n-factor is 5 (it gains 5 electrons).

Let \( V \) be the volume of KMnO4 in liters. The equivalents of KMnO4 can be expressed as:
\[ \text{Equivalents of KMnO4} = \text{Molarity} \times \text{Volume} \times \text{n-factor} \]
\[ \text{Equivalents of KMnO4} = 0.1 \text{ M} \times V \text{ L} \times 5 \]

### Step 4: Set up the equation
Since the equivalents of KMnO4 must equal the equivalents of FeC2O4, we can set up the equation:
\[ 0.1 \times V \times 5 = 0.00208 \]

### Step 5: Solve for V
Now, we can solve for \( V \):
\[ 0.5V = 0.00208 \]
\[ V = \frac{0.00208}{0.5} = 0.00416 \text{ L} \]

### Step 6: Convert volume to milliliters
To convert liters to milliliters:
\[ V = 0.00416 \text{ L} \times 1000 = 4.16 \text{ mL} \]

### Final Answer
The volume of 0.1 M KMnO4 needed to oxidize 100 mg of FeC2O4 in acid solution is approximately **4.16 mL**.

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