if 20 mL `(M)/(10)Ba(MnO_(4))_(2)` compeletely reacts with `FeC_(2)O_(4)` in acidic medium,
Q. What is the volume of `CO_(2)` produced at `STP`

Options

A. 112 mL
B. 224 mL
C. 448 mL
D. none

Explanation

To solve the problem step-by-step, we will follow the stoichiometric calculations based on the reaction of \( \text{Ba(MnO}_4\text{)}_2 \) with \( \text{FeC}_2\text{O}_4 \) in an acidic medium.

### Step 1: Determine the moles of \( \text{Ba(MnO}_4\text{)}_2 \)

Given that the concentration of \( \text{Ba(MnO}_4\text{)}_2 \) is \( \frac{M}{10} \) and the volume is 20 mL, we can calculate the number of moles.

\[ \text{Moles of } \text{Ba(MnO}_4\text{)}_2 = \text{Concentration} \times \text{Volume} = \frac{1}{10} \times 0.020 \, \text{L} = 0.002 \, \text{moles} \]

### Step 2: Determine the equivalent of \( \text{Ba(MnO}_4\text{)}_2 \)

The n-factor for \( \text{Ba(MnO}_4\text{)}_2 \) in acidic medium is 10 (as it reduces \( \text{Mn}^{7+} \) to \( \text{Mn}^{2+} \)). Therefore, the equivalents can be calculated as follows:

\[ \text{Equivalents of } \text{Ba(MnO}_4\text{)}_2 = \text{Moles} \times \text{n-factor} = 0.002 \times 10 = 0.02 \, \text{equivalents} \]

### Step 3: Relate the equivalents to \( \text{FeC}_2\text{O}_4 \)

Since \( \text{FeC}_2\text{O}_4 \) reacts with \( \text{Ba(MnO}_4\text{)}_2 \) in a 1:1 ratio, the equivalents of \( \text{FeC}_2\text{O}_4 \) will also be 0.02 equivalents.

### Step 4: Determine the moles of \( \text{FeC}_2\text{O}_4 \)

The n-factor for \( \text{FeC}_2\text{O}_4 \) is 2 (as each \( \text{C} \) in \( \text{C}_2\text{O}_4^{2-} \) is oxidized from +2 to +4). Therefore, we can find the moles of \( \text{FeC}_2\text{O}_4 \) as follows:

\[ \text{Moles of } \text{FeC}_2\text{O}_4 = \frac{\text{Equivalents}}{\text{n-factor}} = \frac{0.02}{2} = 0.01 \, \text{moles} \]

### Step 5: Calculate the moles of \( \text{CO}_2 \) produced

Each mole of \( \text{FeC}_2\text{O}_4 \) produces 2 moles of \( \text{CO}_2 \). Therefore:

\[ \text{Moles of } \text{CO}_2 = 0.01 \times 2 = 0.02 \, \text{moles} \]

### Step 6: Calculate the volume of \( \text{CO}_2 \) at STP

At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. Therefore, the volume of \( \text{CO}_2 \) produced is:

\[ \text{Volume of } \text{CO}_2 = \text{Moles} \times 22.4 \, \text{L} = 0.02 \times 22.4 = 0.448 \, \text{L} = 448 \, \text{mL} \]

### Final Answer

The volume of \( \text{CO}_2 \) produced at STP is **448 mL**.

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