What is the `("Mol wt")/("Eq.wt")` ratio of `Fe_(2)(SO_(4))_(3)` being converted into `Fe(OH)_(3)`

Answer

To find the ratio of molecular weight to equivalent weight of \( \text{Fe}_2(\text{SO}_4)_3 \) when it is converted into \( \text{Fe(OH)}_3 \), we can follow these steps:

### Step 1: Determine the n-factor (valency factor)
1. **Identify the oxidation states of iron (Fe)**:
- In \( \text{Fe}_2(\text{SO}_4)_3 \), the oxidation state of Fe is +3.
- In \( \text{Fe(OH)}_3 \), the oxidation state of Fe is also +3.

2. **Calculate the n-factor**:
- The n-factor is determined by the change in the number of moles of electrons transferred in the reaction.
- For \( \text{Fe}_2(\text{SO}_4)_3 \), there are 2 moles of Fe, each with a charge of +3, contributing a total of \( 2 \times 3 = 6 \).
- For \( \text{Fe(OH)}_3 \), there is 1 mole of Fe with a charge of +3, contributing a total of \( 1 \times 3 = 3 \).
- Therefore, the n-factor is calculated as:
\[ \text{n-factor} = (6 - 3) = 3 \]

### Step 2: Calculate the molecular weight of \( \text{Fe}_2(\text{SO}_4)_3 \)
1. **Calculate the molecular weight**:
- The molecular weight of \( \text{Fe}_2(\text{SO}_4)_3 \) can be calculated as follows:
- Atomic weight of Fe = 55.85 g/mol
- Atomic weight of S = 32.07 g/mol
- Atomic weight of O = 16.00 g/mol
- Therefore, the molecular weight is:
\[ \text{Molecular weight} = 2 \times 55.85 + 3 \times (32.07 + 4 \times 16.00) \]
\[ = 2 \times 55.85 + 3 \times (32.07 + 64.00) \]
\[ = 111.70 + 3 \times 96.07 = 111.70 + 288.21 = 399.91 \text{ g/mol} \]

### Step 3: Calculate the equivalent weight
1. **Use the formula for equivalent weight**:
- The equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
- Substituting the values:
\[ \text{Equivalent weight} = \frac{399.91}{3} = 133.30 \text{ g/equiv} \]

### Step 4: Calculate the ratio of molecular weight to equivalent weight
1. **Calculate the ratio**:
- The ratio of molecular weight to equivalent weight is:
\[ \text{Ratio} = \frac{\text{Molecular weight}}{\text{Equivalent weight}} = \frac{399.91}{133.30} \approx 3.00 \]

### Final Answer:
The ratio of molecular weight to equivalent weight of \( \text{Fe}_2(\text{SO}_4)_3 \) when converted into \( \text{Fe(OH)}_3 \) is approximately **3.00**.

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