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

Options

A. `(20)/(3)`
B. `(20)/(2)`
C. `(20)/(6)`
D. `(20)/(10)`

Explanation

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

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

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

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

Now, calculate the equivalents:
\[ \text{Equivalents of } Ba(MnO4)2 = \text{Moles} \times \text{n-factor} \]
\[ \text{Equivalents} = 0.002 \, \text{mol} \times 10 = 0.020 \, \text{equivalents} \]

### Step 4: Convert equivalents to milliequivalents
Since 1 equivalent = 1000 milliequivalents:
\[ \text{Milliequivalents of } Ba(MnO4)2 = 0.020 \, \text{equivalents} \times 1000 = 20 \, \text{milliequivalents} \]

### Step 5: Determine the n-factor for `FeC2O4`
In acidic medium, `FeC2O4` oxidizes:
- `Fe^2+` oxidizes to `Fe^3+` (1 electron change)
- `C2O4^2-` oxidizes to `CO2` (2 electrons change)

Total change in oxidation number:
\[ \text{n-factor for } FeC2O4 = 1 + 2 = 3 \]

### Step 6: Calculate millimoles of `FeC2O4`
Using the formula:
\[ \text{Millimoles} = \frac{\text{Milliequivalents}}{\text{n-factor}} \]
\[ \text{Millimoles of } FeC2O4 = \frac{20 \, \text{milliequivalents}}{3} \approx 6.67 \, \text{millimoles} \]

### Final Answer
The millimoles of `FeC2O4` that reacted is approximately **6.67 millimoles**.

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