`FeC_(2)O_(4),Fe_(2)(C_(2)O_(4))_(3),FeSO_(4),Fe_(2)(SO_(4))_(3)` one mole each, will react how many moles of `KMnO_(4)`.

Options

A. 2
B. 1
C. 3
D. `1.5`

Answer

To solve the problem, we need to determine how many moles of KMnO4 will react with one mole each of FeC2O4, Fe2(C2O4)3, FeSO4, and Fe2(SO4)3. We will follow these steps:

### Step 1: Calculate the n-factor of KMnO4
In acidic medium, KMnO4 (MnO4^-) reduces to Mn^2+. The change in oxidation state of Mn is from +7 to +2, which involves a change of 5 electrons.

**n-factor of KMnO4 = 5**

### Step 2: Calculate the n-factor for each iron compound

#### a) For FeC2O4
- Fe is in the +2 oxidation state and can be oxidized to +3. This involves a change of 1 electron.
- The C2O4^2- ion (oxalate) can be oxidized to 2 CO2, which involves a change of 2 electrons.

Total n-factor for FeC2O4:
- n-factor = 1 (for Fe) + 2 (for C2O4) = 3

#### b) For Fe2(C2O4)3
- Fe is already in the +3 oxidation state, so it does not contribute to the n-factor.
- The C2O4^2- ions will still oxidize to CO2, contributing 2 electrons per ion. There are 3 C2O4^2- ions.

Total n-factor for Fe2(C2O4)3:
- n-factor = 2 (for 3 C2O4^2-) = 6

#### c) For FeSO4
- Fe is in the +2 oxidation state and can be oxidized to +3, contributing 1 electron.
- The SO4^2- ion does not change its oxidation state.

Total n-factor for FeSO4:
- n-factor = 1 (for Fe) + 0 (for SO4) = 1

#### d) For Fe2(SO4)3
- Fe is in the +3 oxidation state, so it does not contribute to the n-factor.
- The SO4^2- ions do not change their oxidation state.

Total n-factor for Fe2(SO4)3:
- n-factor = 0 (no contribution from Fe or SO4)

### Step 3: Sum the n-factors
Now we sum the n-factors from each compound:
- FeC2O4: 3
- Fe2(C2O4)3: 6
- FeSO4: 1
- Fe2(SO4)3: 0

Total n-factor = 3 + 6 + 1 + 0 = 10

### Step 4: Calculate the moles of KMnO4 required
Using the relationship between the equivalents of KMnO4 and the total n-factor of iron compounds, we have:

\[ \text{Number of equivalents of KMnO4} = \text{Number of equivalents of Fe} \]

Let \( x \) be the number of moles of KMnO4:
\[ x \times n \text{ (n-factor of KMnO4)} = 1 \text{ (mole of Fe)} \times \text{Total n-factor} \]
\[ x \times 5 = 1 \times 10 \]
\[ x = \frac{10}{5} = 2 \]

### Final Answer
Thus, **2 moles of KMnO4** will react with one mole each of the given iron compounds.

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