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

Answer

To find the molarity of FeCl₃ (aq) when excess NaOH is added, resulting in the formation of 2.14 g of Fe(OH)₃, 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 moles of Fe(OH)₃ formed
To find the moles of Fe(OH)₃, we use the formula:

\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]

First, we need to calculate the molar mass of Fe(OH)₃:
- Molar mass of Fe = 56 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of H = 1 g/mol

Thus, the molar mass of Fe(OH)₃ is:

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

Now, we can calculate the moles of Fe(OH)₃:

\[ \text{Moles of Fe(OH)}_3 = \frac{2.14 \text{ g}}{107 \text{ g/mol}} = 0.02 \text{ mol} \]

### 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 moles of FeCl₃ will also be:

\[ \text{Moles of FeCl}_3 = \text{Moles of Fe(OH)}_3 = 0.02 \text{ mol} \]

### Step 4: Calculate the molarity of FeCl₃
Molarity (M) is defined as moles of solute per liter of solution. The volume of the FeCl₃ solution is given as 100 mL, which we need to convert to liters:

\[ 100 \text{ mL} = 0.1 \text{ L} \]

Now we can calculate the molarity:

\[ \text{Molarity of FeCl}_3 = \frac{\text{Moles of FeCl}_3}{\text{Volume in L}} = \frac{0.02 \text{ mol}}{0.1 \text{ L}} = 0.2 \text{ M} \]

### Final Answer
The molarity of FeCl₃ (aq) is **0.2 M**.

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