A solution containing 0.2 mole of ferric chloride is allowed to react with 0.24 mole of sodium hydroxide. The correct statement for this reaction is

Options

A. Limiting reagent for this reaction is `NaOH`
B. 0.08 moles of `Fe(OH)_(3)` is formed
C. 0.12 mole of `FeCl_(3)` is left unreacted
D. All the above

Explanation

To determine the correct statement regarding the reaction between ferric chloride (FeCl3) and sodium hydroxide (NaOH), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction between ferric chloride and sodium hydroxide can be represented as follows:

\[ \text{FeCl}_3 + 3 \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \text{NaCl} \]

### Step 2: Identify the Moles of Reactants
From the question:
- Moles of ferric chloride (FeCl3) = 0.2 moles
- Moles of sodium hydroxide (NaOH) = 0.24 moles

### Step 3: Determine the Stoichiometry
According to the balanced equation, 1 mole of FeCl3 reacts with 3 moles of NaOH. Therefore, to react with 0.2 moles of FeCl3, we need:

\[ 0.2 \, \text{moles of FeCl}_3 \times 3 \, \text{moles of NaOH/mole of FeCl}_3 = 0.6 \, \text{moles of NaOH} \]

### Step 4: Identify the Limiting Reagent
We have 0.24 moles of NaOH available, but we need 0.6 moles to completely react with 0.2 moles of FeCl3. Since we do not have enough NaOH, it is the limiting reagent in this reaction.

### Step 5: Calculate the Amount of Fe(OH)3 Produced
Since NaOH is the limiting reagent, we can calculate how much Fe(OH)3 will be produced from the available NaOH:

From the balanced equation, 3 moles of NaOH produce 1 mole of Fe(OH)3. Therefore, the moles of Fe(OH)3 produced from 0.24 moles of NaOH is:

\[ \frac{0.24 \, \text{moles of NaOH}}{3} = 0.08 \, \text{moles of Fe(OH)}_3 \]

### Conclusion
Based on the calculations, the correct statement regarding the reaction is that sodium hydroxide (NaOH) is the limiting reagent, and it will produce 0.08 moles of Fe(OH)3.