21 Mol of `FeSO_(2)` (atomic weight of Fe is 55.84 g `mol^(-1))` is oxidized to `Fe_(2)(SO_(4))^(3)` calculate the equivalent weight of ferrous ion

Options

A. 55.84
B. 27.92
C. 18.61
D. 111.68

Explanation

To calculate the equivalent weight of ferrous ion (Fe²⁺) when it is oxidized to ferric ion (Fe³⁺), we can follow these steps:

### Step 1: Identify the oxidation process
When ferrous ion (Fe²⁺) is oxidized to ferric ion (Fe³⁺), it loses one electron:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

### Step 2: Determine the molecular weight of ferrous ion
The atomic weight of iron (Fe) is given as 55.84 g/mol. Therefore, the molecular weight of ferrous ion (Fe²⁺) is also 55.84 g/mol.

### Step 3: Determine the number of electrons transferred (n)
In this reaction, each ferrous ion (Fe²⁺) loses one electron to become ferric ion (Fe³⁺). Thus, the value of n (the number of electrons transferred) is 1.

### Step 4: Calculate the equivalent weight
The equivalent weight (E) can be calculated using the formula:
\[ E = \frac{\text{Molecular Weight}}{n} \]
Substituting the values:
\[ E = \frac{55.84 \, \text{g/mol}}{1} = 55.84 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of ferrous ion (Fe²⁺) is **55.84 g/equiv**.

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