Excess of NaOH (aq) was added to 100 mL of `FeCI_(3)` (aq) resulting into 2.14 g of `Fe(OH)_(3).` The molarity of `FeCI_(3)` (aq) is:
(Given molar mass of Fe = 56g `mol^(-1)` and molar mass of CI = 35.5g `mol^(-1)`)

Options

A. 0.2 M
B. 1.8 M
C. 0.3 M
D. 0.6 M

Explanation

To find the molarity of `FeCl3` in the solution, we can follow these steps:

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

### Step 2: Calculate the molar mass of `Fe(OH)3`
To find the molar mass of `Fe(OH)3`, we add the molar masses of its constituent elements:
- Molar mass of Fe = 56 g/mol
- Molar mass of O = 16 g/mol (3 O atoms)
- Molar mass of H = 1 g/mol (3 H atoms)

Calculating the total:
\[ \text{Molar mass of Fe(OH)}_3 = 56 + (3 \times 16) + (3 \times 1) = 56 + 48 + 3 = 107 \text{ g/mol} \]

### Step 3: Calculate the moles of `Fe(OH)3`
Using the mass of `Fe(OH)3` produced (2.14 g), we can calculate the number of moles:
\[ \text{Moles of Fe(OH)}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.14 \text{ g}}{107 \text{ g/mol}} \approx 0.02 \text{ moles} \]

### Step 4: 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 moles of `FeCl3` are also 0.02 moles.

### Step 5: Calculate the molarity of `FeCl3`
Molarity (M) is defined as the number of moles of solute per liter of solution. We have 0.02 moles of `FeCl3` in 100 mL of solution. First, convert the volume from mL to L:
\[ 100 \text{ mL} = 0.1 \text{ L} \]

Now, we can calculate the molarity:
\[ \text{Molarity of FeCl}_3 = \frac{\text{moles of FeCl}_3}{\text{volume in L}} = \frac{0.02 \text{ moles}}{0.1 \text{ L}} = 0.2 \text{ M} \]

### Final Answer
The molarity of `FeCl3` is **0.2 M**.

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