The equivalent weight of `FeS_(2)` in the following reaction is `FeS_(2)+O_(2) rarr Fe^(+3)+SO_(2)`

Options

A. `"Mol. wt"/1`
B. `"Mol. wt"/7`
C. `"Mol. wt"/11`
D. `"Mol. wt"/9`

Answer

To find the equivalent weight of \( \text{FeS}_2 \) in the reaction:

\[ \text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}^{3+} + \text{SO}_2 \]

we will follow these steps:

### Step 1: Write the balanced equation
First, we need to balance the reaction. The unbalanced equation is:

\[ \text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}^{3+} + \text{SO}_2 \]

Balancing the equation, we find:

\[ 4 \text{FeS}_2 + 11 \text{O}_2 \rightarrow 4 \text{Fe}^{3+} + 8 \text{SO}_2 \]

### Step 2: Determine the number of electrons transferred
In this reaction, each \( \text{FeS}_2 \) molecule contributes to the oxidation of iron from \( \text{Fe}^{0} \) in \( \text{FeS}_2 \) to \( \text{Fe}^{3+} \). Each iron atom loses 3 electrons. Therefore, for 4 moles of \( \text{FeS}_2 \):

\[ \text{Total electrons lost} = 4 \text{ moles of Fe} \times 3 \text{ electrons} = 12 \text{ electrons} \]

Additionally, sulfur in \( \text{FeS}_2 \) is oxidized to \( \text{SO}_2 \), which involves a change in oxidation state from -1 (in \( \text{FeS}_2 \)) to +4 (in \( \text{SO}_2 \)). Each sulfur atom loses 5 electrons, and since there are 8 sulfur atoms in 4 moles of \( \text{FeS}_2 \):

\[ \text{Total electrons lost by sulfur} = 8 \text{ moles of S} \times 5 \text{ electrons} = 40 \text{ electrons} \]

Thus, the total number of electrons transferred in the reaction is:

\[ \text{Total electrons} = 12 + 40 = 52 \text{ electrons} \]

### Step 3: Calculate the n-factor
The n-factor for \( \text{FeS}_2 \) is the total number of electrons transferred per mole of \( \text{FeS}_2 \) in the balanced reaction. Since we have 4 moles of \( \text{FeS}_2 \) transferring 52 electrons:

\[ \text{n-factor} = \frac{52 \text{ electrons}}{4 \text{ moles of FeS}_2} = 13 \]

### Step 4: Find the molar mass of \( \text{FeS}_2 \)
Next, we calculate the molar mass of \( \text{FeS}_2 \):

- Atomic mass of Fe = 55.85 g/mol
- Atomic mass of S = 32.07 g/mol

Thus, the molar mass of \( \text{FeS}_2 \) is:

\[ \text{Molar mass of } \text{FeS}_2 = 55.85 + 2 \times 32.07 = 119.89 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight
Finally, we can calculate the equivalent weight using the formula:

\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{119.89 \text{ g/mol}}{13} \approx 9.23 \text{ g/equiv} \]

### Conclusion
The equivalent weight of \( \text{FeS}_2 \) in the given reaction is approximately 9.23 g/equiv.

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