In the redox reaction -
`10FeC_(2) O_(4)+x KMnO_(4)+24H_(2)SO_(4) rarr 5Fe_(2) ( SO_(4))_(3) + 20 CO_(2) + y MnSO_(4) + 3K_(2) SO_(4)+ 24 H_(2)O`
The value of x and y are respectively -

Options

A. 6.3
B. 3.6
C. 3.3
D. 6.6

Explanation

To determine the values of \( x \) and \( y \) in the redox reaction:

\[ 10 \text{FeC}_2\text{O}_4 + x \text{KMnO}_4 + 24 \text{H}_2\text{SO}_4 \rightarrow 5 \text{Fe}_2(\text{SO}_4)_3 + 20 \text{CO}_2 + y \text{MnSO}_4 + 3 \text{K}_2\text{SO}_4 + 24 \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Determine Oxidation States
- **Iron (Fe)** in FeC₂O₄ is in the +2 oxidation state.
- **Carbon (C)** in oxalate (C₂O₄²⁻) can be calculated as follows:
- Let the oxidation state of C be \( x \).
- The equation becomes: \( 2x + 4(-2) = -2 \) (total charge).
- Solving gives \( 2x - 8 = -2 \) → \( 2x = 6 \) → \( x = +3 \).
- **Manganese (Mn)** in KMnO₄ is in the +7 oxidation state.
- **Sulfur (S)** in H₂SO₄ is in the +6 oxidation state.

### Step 2: Identify Changes in Oxidation States
- **Oxidation**:
- Fe²⁺ (from FeC₂O₄) is oxidized to Fe³⁺.
- Each Fe²⁺ loses 1 electron, and since there are 10 Fe²⁺, total electrons lost = 10.

- **Reduction**:
- MnO₄⁻ (Mn in +7) is reduced to Mn²⁺ (Mn in +2).
- Each Mn gains 5 electrons.

### Step 3: Balance the Electrons
To balance the electrons lost and gained:
- 10 electrons are lost by Fe.
- Let \( x \) be the number of KMnO₄ used. Each KMnO₄ gains 5 electrons, so \( 5x \) electrons are gained.

Setting the electrons equal:
\[ 10 = 5x \implies x = 2 \]

### Step 4: Determine the Value of \( y \)
From the balanced reaction, each KMnO₄ produces 1 MnSO₄. Therefore, if \( x = 2 \), then \( y = 2 \).

### Final Balanced Reaction
The balanced equation becomes:
\[ 10 \text{FeC}_2\text{O}_4 + 2 \text{KMnO}_4 + 24 \text{H}_2\text{SO}_4 \rightarrow 5 \text{Fe}_2(\text{SO}_4)_3 + 20 \text{CO}_2 + 2 \text{MnSO}_4 + 3 \text{K}_2\text{SO}_4 + 24 \text{H}_2\text{O} \]

### Conclusion
Thus, the values of \( x \) and \( y \) are:
\[ x = 2 \quad \text{and} \quad y = 2 \]