An excess of NaOH was added to 100 mL of a ferric chloride solution.This caused the precipitation of 1.425 g of `Fe(OH)_3`.Calculate the normality of the ferric chloride solution.

Options

A. 0.20 N
B. 0.50 N
C. 0.25 N
D. 0.40 N

Explanation

To solve the problem of calculating the normality of the ferric chloride solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between ferric chloride (FeCl₃) and sodium hydroxide (NaOH) can be represented as:
\[ \text{FeCl}_3 + 3 \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \text{NaCl} \]

### Step 2: Calculate the number of moles of Fe(OH)₃
Given that 1.425 g of Fe(OH)₃ is precipitated, we first need to calculate the number of moles of Fe(OH)₃ formed.

The molar mass of Fe(OH)₃ is:
- Iron (Fe): 55.85 g/mol
- Oxygen (O): 16.00 g/mol (3 O atoms = 48.00 g/mol)
- Hydrogen (H): 1.01 g/mol (3 H atoms = 3.03 g/mol)

Total molar mass of Fe(OH)₃:
\[ 55.85 + 48.00 + 3.03 = 106.88 \, \text{g/mol} \]

Now, we calculate the number of moles of Fe(OH)₃:
\[ \text{Number of moles of Fe(OH)}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.425 \, \text{g}}{106.88 \, \text{g/mol}} \approx 0.0133 \, \text{mol} \]

### Step 3: Relate moles of Fe(OH)₃ to moles of FeCl₃
From the balanced equation, we see that 1 mole of FeCl₃ produces 1 mole of Fe(OH)₃. Therefore, the number of moles of FeCl₃ is also:
\[ \text{Number of moles of FeCl}_3 = 0.0133 \, \text{mol} \]

### Step 4: Calculate the gram equivalent weight of FeCl₃
The equivalent weight of FeCl₃ can be calculated using its molar mass and the n-factor. The n-factor for FeCl₃ is 3 (since it can donate 3 moles of Cl⁻ ions).

The molar mass of FeCl₃ is:
- Iron (Fe): 55.85 g/mol
- Chlorine (Cl): 35.45 g/mol (3 Cl atoms = 106.35 g/mol)

Total molar mass of FeCl₃:
\[ 55.85 + 106.35 = 162.20 \, \text{g/mol} \]

Now, calculate the gram equivalent weight:
\[ \text{Gram equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{162.20 \, \text{g/mol}}{3} \approx 54.07 \, \text{g/equiv} \]

### Step 5: Calculate the given weight of FeCl₃
Using the number of moles of FeCl₃, we can find the given weight:
\[ \text{Given weight of FeCl}_3 = \text{Number of moles} \times \text{Molar mass} = 0.0133 \, \text{mol} \times 162.20 \, \text{g/mol} \approx 2.16 \, \text{g} \]

### Step 6: Calculate the number of gram equivalents of FeCl₃
Now, we can find the number of gram equivalents of FeCl₃:
\[ \text{Number of gram equivalents} = \frac{\text{Given weight}}{\text{Gram equivalent weight}} = \frac{2.16 \, \text{g}}{54.07 \, \text{g/equiv}} \approx 0.04 \, \text{equiv} \]

### Step 7: Calculate the normality of the ferric chloride solution
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. Since we have 100 mL of solution, we convert it to liters:
\[ \text{Volume in liters} = \frac{100 \, \text{mL}}{1000} = 0.1 \, \text{L} \]

Now, we can calculate the normality:
\[ N = \frac{\text{Number of gram equivalents}}{\text{Volume in L}} = \frac{0.04 \, \text{equiv}}{0.1 \, \text{L}} = 0.4 \, \text{N} \]

### Final Answer
The normality of the ferric chloride solution is **0.4 N**.

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