Five moles of ferric oxalate are oxidise by how much mole of `KMnO_(4)` in acidic medium.

Explanation

To solve the problem of how many moles of KMnO4 are required to oxidize 5 moles of ferric oxalate in acidic medium, we will follow these steps:

### Step 1: Identify the chemical formulas
Ferric oxalate is represented as Fe2(C2O4)3. In this case, we have 5 moles of ferric oxalate.

### Step 2: Determine the oxidation states
In acidic medium, KMnO4 (potassium permanganate) is reduced from Mn in +7 oxidation state to Mn in +2 oxidation state. The change in oxidation state indicates that each mole of KMnO4 can accept 5 electrons (n-factor = 5).

### Step 3: Calculate the n-factor for ferric oxalate
For ferric oxalate (Fe2(C2O4)3), the iron (Fe) has an oxidation state of +3. Since there are 2 Fe atoms, the total charge contributed by Fe is:
- Charge from Fe = 2 atoms × +3 = +6.

The oxalate ion (C2O4) has a charge of -2, and since there are 3 oxalate ions in ferric oxalate, the total charge from oxalate is:
- Charge from C2O4 = 3 ions × -2 = -6.

The total n-factor for ferric oxalate is the total number of electrons transferred, which is 6.

### Step 4: Set up the equivalence equation
Using the concept of equivalents, we can set up the equation:
\[ \text{Equivalents of Fe2(C2O4)3} = \text{Equivalents of KMnO4} \]
Where:
\[ \text{Equivalents} = \text{n-factor} \times \text{number of moles} \]

For ferric oxalate:
\[ \text{Equivalents of Fe2(C2O4)3} = 6 \times 5 = 30 \]

For KMnO4:
\[ \text{Equivalents of KMnO4} = 5 \times \text{number of moles of KMnO4} \]

### Step 5: Equate the equivalents
Setting the equivalents equal gives us:
\[ 30 = 5 \times \text{number of moles of KMnO4} \]

### Step 6: Solve for the number of moles of KMnO4
To find the number of moles of KMnO4, we rearrange the equation:
\[ \text{number of moles of KMnO4} = \frac{30}{5} = 6 \]

### Final Answer
Thus, 6 moles of KMnO4 are required to oxidize 5 moles of ferric oxalate in acidic medium.

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