if 20 mL `(M)/(10)Ba(MnO_(4))_(2)` compeletely reacts with `FeC_(2)O_(4)` in acidic medium,
Q. m" Eq of "`FeC_(2)O_(4)` reacted is

Options

A. 6
B. 20
C. 40
D. none

Explanation

To solve the problem, we need to determine the milliequivalents of `FeC2O4` that reacted with `Ba(MnO4)2` in an acidic medium. Here’s a step-by-step solution:

### Step 1: Determine the Molarity of `Ba(MnO4)2`
Given that the concentration of `Ba(MnO4)2` is `M/10`, we can express this as:
- Molarity (M) = 0.1 M

### Step 2: Calculate the Number of Moles of `Ba(MnO4)2`
Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]
Convert 20 mL to liters:
\[ 20 \text{ mL} = 0.020 \text{ L} \]
Now calculate the moles:
\[ \text{Moles of } Ba(MnO4)2 = 0.1 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{moles} \]

### Step 3: Determine the n-factor for `Ba(MnO4)2`
In acidic medium, `MnO4^-` is reduced from an oxidation state of +7 to +2, which involves a change of 5 electrons per `MnO4^-`. Since there are 2 `MnO4^-` ions in `Ba(MnO4)2`, the total change in electrons is:
\[ \text{n-factor of } Ba(MnO4)2 = 2 \times 5 = 10 \]

### Step 4: Calculate the Total Equivalents of `Ba(MnO4)2`
Using the formula:
\[ \text{Equivalents} = \text{Moles} \times \text{n-factor} \]
\[ \text{Equivalents of } Ba(MnO4)2 = 0.002 \, \text{moles} \times 10 = 0.02 \, \text{equivalents} \]
Convert to milliequivalents:
\[ 0.02 \, \text{equivalents} = 20 \, \text{milliequivalents} \]

### Step 5: Determine the n-factor for `FeC2O4`
In the reaction, `FeC2O4` is oxidized. The oxidation states change as follows:
- `Fe` is oxidized from +2 to +3 (1 electron)
- `C2O4^2-` is oxidized to `CO2` (2 electrons)

Thus, the total change in electrons (n-factor) for `FeC2O4` is:
\[ \text{n-factor of } FeC2O4 = 1 + 2 = 3 \]

### Step 6: Calculate the Milliequivalents of `FeC2O4`
Since the equivalents of `Ba(MnO4)2` must equal the equivalents of `FeC2O4` that reacted:
\[ \text{Equivalents of } FeC2O4 = 20 \, \text{milliequivalents} \]

### Final Answer
The milliequivalents of `FeC2O4` that reacted is **20 mEq**.

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