1 mole of equimolar mixture of ferric oxalate and ferrous oxalate requres x mole of `KMnO_(4)` in acidic medium for complete oxidation. X is:

Options

A. 0.5 mole
B. 0.9 mole
C. 1.2 mole
D. 4.5 mole

Answer

To solve the problem of how many moles of KMnO₄ are required to completely oxidize 1 mole of an equimolar mixture of ferric oxalate and ferrous oxalate in acidic medium, we can follow these steps:

### Step 1: Understand the Composition of the Mixture
The mixture consists of:
- 0.5 moles of ferric oxalate (Fe₂(C₂O₄)₃)
- 0.5 moles of ferrous oxalate (FeC₂O₄)

### Step 2: Determine the Oxidation States
- In ferric oxalate (Fe₂(C₂O₄)₃), iron is in the +3 oxidation state.
- In ferrous oxalate (FeC₂O₄), iron is in the +2 oxidation state.

### Step 3: Write the Oxidation Reactions
1. **For Ferric Oxalate (Fe₂(C₂O₄)₃)**:
- The complete oxidation of ferric oxalate can be represented as:
\[ Fe_2(C_2O_4)_3 + 6 KMnO_4 + 12 H_2SO_4 \rightarrow 2 Fe_2(SO_4)_3 + 6 MnSO_4 + 12 CO_2 + 12 H_2O \]
- From this reaction, we see that 1 mole of ferric oxalate requires 3 moles of KMnO₄.

2. **For Ferrous Oxalate (FeC₂O₄)**:
- The complete oxidation of ferrous oxalate can be represented as:
\[ FeC_2O_4 + KMnO_4 + H_2SO_4 \rightarrow Fe_2(SO_4)_3 + MnSO_4 + 2 CO_2 + 2 H_2O \]
- From this reaction, we see that 1 mole of ferrous oxalate requires 1 mole of KMnO₄.

### Step 4: Calculate Total KMnO₄ Required
- For 0.5 moles of ferric oxalate:
\[ 0.5 \text{ moles} \times 3 \text{ moles of KMnO₄} = 1.5 \text{ moles of KMnO₄} \]
- For 0.5 moles of ferrous oxalate:
\[ 0.5 \text{ moles} \times 1 \text{ mole of KMnO₄} = 0.5 \text{ moles of KMnO₄} \]

### Step 5: Add the Moles of KMnO₄
- Total moles of KMnO₄ required:
\[ 1.5 \text{ moles} + 0.5 \text{ moles} = 2 \text{ moles of KMnO₄} \]

### Final Answer
Thus, the value of x is:
\[ \boxed{2} \]

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