Dichloroacetic acid `(CHCI_2COOH)` is oxidised to `CO_(2),H_(2)O` and `CI_(2)` by 600meq of an oxidising agent. Same amount of ammonia to form ammonium dichloroacetate:

Options

A. 0.0167
B. 0.1
C. 0.3
D. 0.6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the oxidation reaction of dichloroacetic acid
Dichloroacetic acid (CHCl₂COOH) is oxidized to carbon dioxide (CO₂), chlorine gas (Cl₂), and water (H₂O). The balanced reaction can be represented as:
\[ \text{CHCl}_2\text{COOH} \rightarrow 2 \text{CO}_2 + \text{Cl}_2 + \text{H}_2\text{O} \]

### Step 2: Determine the equivalents of dichloroacetic acid
From the reaction, we can see that 1 mole of dichloroacetic acid produces 1 equivalent of products. Therefore, we can relate the equivalents of the oxidizing agent to the moles of dichloroacetic acid.

### Step 3: Convert the given milliequivalents to equivalents
We are given that 600 milliequivalents (meq) of the oxidizing agent are used. To convert this to equivalents:
\[ 600 \text{ meq} = 600 \times 10^{-3} \text{ eq} = 0.6 \text{ eq} \]

### Step 4: Calculate moles of dichloroacetic acid
According to the stoichiometry of the reaction, 1 equivalent of dichloroacetic acid corresponds to 1/6 moles (since 1 mole produces 6 equivalents). Therefore, we can calculate the moles of dichloroacetic acid required:
\[ \text{Moles of CHCl}_2\text{COOH} = \frac{0.6 \text{ eq}}{6} = 0.1 \text{ moles} \]

### Step 5: Write the reaction of ammonia with dichloroacetic acid
When ammonia (NH₃) reacts with dichloroacetic acid, it forms ammonium dichloroacetate:
\[ \text{CHCl}_2\text{COOH} + \text{NH}_3 \rightarrow \text{NH}_4\text{CHCl}_2\text{COO} \]

### Step 6: Determine the amount of ammonia required
Since 1 mole of dichloroacetic acid reacts with 1 mole of ammonia, the moles of ammonia required will also be equal to the moles of dichloroacetic acid:
\[ \text{Moles of NH}_3 = 0.1 \text{ moles} \]

### Conclusion
Thus, the amount of ammonia required to form ammonium dichloroacetate is **0.1 moles**.

---