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: Determine the oxidation states of the elements in \( \text{FeS}_2 \)

In \( \text{FeS}_2 \):
- Iron (Fe) is in the +2 oxidation state.
- Sulfur (S) is in the -1 oxidation state (since there are two sulfur atoms, the total contribution is -2, balancing the +2 from iron).

### Step 2: Identify the oxidation states after the reaction

In the products:
- Iron (Fe) is oxidized to +3.
- Sulfur (S) in \( \text{SO}_2 \) is in the +4 oxidation state.

### Step 3: Calculate the change in oxidation states

For iron:
- Change in oxidation state = \( +3 - (+2) = +1 \)

For sulfur:
- Change in oxidation state = \( +4 - (-1) = +5 \)

### Step 4: Calculate the total change in oxidation states

Since there are two sulfur atoms in \( \text{FeS}_2 \), the total change for sulfur will be:
- Total change for sulfur = \( 2 \times 5 = 10 \)

Thus, the total change in oxidation states (N) for the reaction is:
- Total change (N) = Change for Fe + Change for S = \( 1 + 10 = 11 \)

### Step 5: Calculate the molecular weight of \( \text{FeS}_2 \)

The molecular weight of \( \text{FeS}_2 \) can be calculated as follows:
- Atomic weight of Fe = 55.85 g/mol
- Atomic weight of S = 32.07 g/mol

So, the molecular weight of \( \text{FeS}_2 \) is:
\[ \text{Molecular weight of } \text{FeS}_2 = 55.85 + 2 \times 32.07 = 55.85 + 64.14 = 119.99 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight

The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{N} \]
Substituting the values we have:
\[ \text{Equivalent weight} = \frac{119.99 \text{ g/mol}}{11} \approx 10.91 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( \text{FeS}_2 \) in the given reaction is approximately \( 10.91 \text{ g/equiv} \).

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