The equivalent weight of `FeC_(2)O_(4)` in the change
`FeC_(2)O_(4)rarrFe^(3+)+CO_(2)` is

Options

A. `M//3`
B. `M//6`
C. `M//2`
D. `M//1`

Explanation

To find the equivalent weight of `FeC_(2)O_(4)` in the reaction:

\[ \text{FeC}_2\text{O}_4 \rightarrow \text{Fe}^{3+} + \text{CO}_2 \]

we will follow these steps:

### Step 1: Identify the oxidation states
In `FeC_(2)O_(4)`, iron (Fe) has an oxidation state of +2, and each carbon in `C_(2)O_(4)` has an oxidation state of +3. The overall charge of the compound is neutral.

### Step 2: Determine the changes in oxidation states
In the reaction, iron is oxidized from +2 to +3. Each carbon in `C_(2)O_(4)` is oxidized to `CO2`, where carbon has an oxidation state of +4.

### Step 3: Calculate the total number of electrons transferred
- For iron:
- Change: +2 to +3 = 1 electron lost (oxidation).

- For carbon:
- Each carbon changes from +3 to +4 = 1 electron lost per carbon.
- Since there are 2 carbons, total = 2 electrons lost (oxidation).

Thus, the total number of electrons transferred in the reaction is:
\[ 1 \text{ (from Fe)} + 2 \text{ (from 2 C)} = 3 \text{ electrons} \]

### Step 4: Define the n-factor
The n-factor is defined as the total number of electrons transferred in the reaction. Therefore, the n-factor for `FeC_(2)O_(4)` in this reaction is 3.

### Step 5: Calculate the molar mass of `FeC_(2)O_(4)`
To calculate the equivalent weight, we first need the molar mass of `FeC_(2)O_(4)`:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of C = 12.01 g/mol (2 carbons = 2 × 12.01 = 24.02 g/mol)
- Molar mass of O = 16.00 g/mol (4 oxygens = 4 × 16.00 = 64.00 g/mol)

Adding these together:
\[ \text{Molar mass of } FeC_2O_4 = 55.85 + 24.02 + 64.00 = 143.87 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight
The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n-factor}} \]

Substituting the values:
\[ \text{Equivalent Weight} = \frac{143.87 \text{ g/mol}}{3} = 47.96 \text{ g/equiv} \]

### Final Answer:
The equivalent weight of `FeC_(2)O_(4)` in the given reaction is approximately **47.96 g/equiv**.

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