A sample of `FeSO_(4)` and `FeC_(2)O_(4)` is dissolved in `H_(2)SO_(4)` . The complete oxidation of sample required 8/3 eq. of `KMnO_(4)`. After oxidation , the reaction mixture was reduced by Z. On again oxidation by `KMnO_(4) ` required `5/3` eq. The mole ratio of `FeSO_(4)` and `FeC_(2)O_(4)` is

Options

A. `3/7`
B. `7/3`
C. `5/7`
D. `7/5`

Answer

To find the mole ratio of `FeSO4` and `FeC2O4` in the given reaction, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = moles of `FeSO4`
- \( y \) = moles of `FeC2O4`

### Step 2: Determine the Reaction with KMnO4
The complete oxidation of the sample requires \( \frac{8}{3} \) equivalents of `KMnO4`. The oxidation reactions are as follows:
- `FeSO4` (Fe²⁺) is oxidized to `Fe³⁺`, which requires 1 equivalent.
- `FeC2O4` (Fe²⁺) is oxidized to `Fe³⁺`, and the `C2O4²⁻` is oxidized to `2CO2`, which requires 2 equivalents (1 for Fe and 1 for C2O4).

### Step 3: Write the Equation for the First Oxidation
The total equivalents can be expressed as:
\[ x \cdot 1 + y \cdot 2 = \frac{8}{3} \]
This simplifies to:
\[ x + 2y = \frac{8}{3} \quad \text{(Equation 1)} \]

### Step 4: Determine the Reaction after Reduction by Zinc
After reduction by zinc, the mixture is oxidized again by `KMnO4`, requiring \( \frac{5}{3} \) equivalents. The reactions are:
- `FeSO4` (Fe²⁺) remains as `Fe²⁺` after reduction.
- `FeC2O4` (Fe²⁺) remains as `Fe²⁺` after reduction, but the `C2O4²⁻` is oxidized to `2CO2`, which requires 2 equivalents.

### Step 5: Write the Equation for the Second Oxidation
The total equivalents for this oxidation can be expressed as:
\[ x \cdot 1 + y \cdot 1 = \frac{5}{3} \]
This simplifies to:
\[ x + y = \frac{5}{3} \quad \text{(Equation 2)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( x + 2y = \frac{8}{3} \)
2. \( x + y = \frac{5}{3} \)

We can solve these equations simultaneously.

From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = \frac{5}{3} - y \]

Substituting this expression for \( x \) into Equation 1:
\[ \left(\frac{5}{3} - y\right) + 2y = \frac{8}{3} \]
\[ \frac{5}{3} + y = \frac{8}{3} \]
\[ y = \frac{8}{3} - \frac{5}{3} = 1 \]

Now substituting \( y = 1 \) back into Equation 2 to find \( x \):
\[ x + 1 = \frac{5}{3} \]
\[ x = \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]

### Step 7: Calculate the Mole Ratio
Now we have:
- \( x = \frac{2}{3} \) (moles of `FeSO4`)
- \( y = 1 \) (moles of `FeC2O4`)

The mole ratio of `FeSO4` to `FeC2O4` is:
\[ \text{Mole Ratio} = \frac{x}{y} = \frac{\frac{2}{3}}{1} = \frac{2}{3} \]

### Final Answer
Thus, the mole ratio of `FeSO4` to `FeC2O4` is \( \frac{2}{3} \).

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