What is `("mol. Wt")/("Eq. wt")` of `FeC_(2)O_(4)` getting converted into `Fe^(+3)` and `CO_(2)`?

Answer

To solve the problem of finding the ratio of molecular weight to equivalent weight of `FeC₂O₄` (ferrous oxalate) when it gets converted into `Fe^(+3)` and `CO₂`, we can follow these steps:

### Step 1: Determine the Molecular Weight of `FeC₂O₄`
The molecular weight of `FeC₂O₄` can be calculated by adding the atomic weights of its constituent elements:
- Iron (Fe): 55.85 g/mol
- Carbon (C): 12.01 g/mol (2 atoms)
- Oxygen (O): 16.00 g/mol (4 atoms)

**Calculation:**
\[ \text{Molecular Weight} = 55.85 + (2 \times 12.01) + (4 \times 16.00) \]
\[ = 55.85 + 24.02 + 64.00 = 143.87 \text{ g/mol} \]

### Step 2: Identify the Change in Oxidation States
In the reaction:
- `FeC₂O₄` (where Fe is in +2 oxidation state) is converted to `Fe^(+3)` (where Fe is in +3 oxidation state).
- The carbon in `C₂O₄^(2-)` has an average oxidation state of +3 and is converted to `CO₂` where carbon has an oxidation state of +4.

### Step 3: Calculate the n-factor
The n-factor is defined as the total number of moles of electrons exchanged in the reaction.

1. **For Iron (Fe):**
- Change from +2 to +3:
\[ \text{Change} = 3 - 2 = 1 \text{ electron} \]

2. **For Carbon (C):**
- Each carbon changes from +3 to +4:
\[ \text{Change per carbon} = 4 - 3 = 1 \text{ electron} \]
- Since there are 2 carbon atoms:
\[ \text{Total change for carbon} = 2 \times 1 = 2 \text{ electrons} \]

3. **Total n-factor:**
\[ \text{Total n-factor} = 1 \text{ (from Fe)} + 2 \text{ (from C)} = 3 \]

### Step 4: Calculate the Equivalent Weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]

### Step 5: Find the Ratio of Molecular Weight to Equivalent Weight
Using the relationship:
\[ \frac{\text{Molecular Weight}}{\text{Equivalent Weight}} = \text{n-factor} \]

From our calculations, we found that the n-factor is 3. Therefore:
\[ \frac{\text{Molecular Weight}}{\text{Equivalent Weight}} = 3 \]

### Final Answer
The ratio of molecular weight to equivalent weight of `FeC₂O₄` is **3**.

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