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 `FeC₂O₄` 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 `FeC₂O₄`, iron (Fe) is in the +2 oxidation state, and the oxalate ion (C₂O₄) has a -2 charge overall.

### Step 2: Write the half-reactions
The reaction can be split into two half-reactions:

1. Oxidation half-reaction:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
Here, one electron is lost when `Fe²⁺` is oxidized to `Fe³⁺`.

2. Reduction half-reaction:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 e^- \]
In this case, the oxalate ion is oxidized to carbon dioxide, releasing two electrons.

### Step 3: Combine the half-reactions
To balance the overall reaction, we need to ensure that the number of electrons lost equals the number of electrons gained. The total number of electrons involved in the reaction is 3 (1 from Fe and 2 from C₂O₄).

Thus, we can write:
\[ \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Fe}^{3+} + 2 \text{CO}_2 + 2 e^- \]

### Step 4: Calculate the molecular weight of `FeC₂O₄`
The molecular weight of `FeC₂O₄` can be calculated as follows:
- Atomic weight of Fe = 55.85 g/mol
- Atomic weight of C = 12.01 g/mol (2 atoms)
- Atomic weight of O = 16.00 g/mol (4 atoms)

So, the molecular weight (M) of `FeC₂O₄` is:
\[ M = 55.85 + (2 \times 12.01) + (4 \times 16.00) = 55.85 + 24.02 + 64.00 = 143.87 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight
The equivalent weight is defined as:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Number of moles of electrons transferred}} \]
From our half-reactions, we determined that 3 moles of electrons are involved in the reaction.

Thus, the equivalent weight of `FeC₂O₄` is:
\[ \text{Equivalent Weight} = \frac{143.87 \text{ g/mol}}{3} \approx 47.96 \text{ g/equiv} \]

### Final Answer
The equivalent weight of `FeC₂O₄` in the given reaction is approximately **47.96 g/equiv**.

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