How many moles of AgCl would be obtained, when `Co(NH_(3) )_(5)Cl_(3)` is treated with excess of `AgNO_(3)`?

Options

A. 1 mol
B. 2 mol
C. 3 mol
D. No ppt is formed

Explanation

To determine how many moles of AgCl would be obtained when Co(NH₃)₅Cl₃ is treated with excess AgNO₃, we can follow these steps:

### Step 1: Understand the Complex
The complex Co(NH₃)₅Cl₃ consists of a cobalt ion (Co) surrounded by five ammonia (NH₃) ligands and three chloride (Cl) ligands.

### Step 2: Dissociation of the Complex
When Co(NH₃)₅Cl₃ dissociates, it can be represented as:
\[ \text{Co(NH}_3\text{)}_5\text{Cl}_3 \rightarrow \text{Co(NH}_3\text{)}_5\text{Cl}^{2+} + 2\text{Cl}^- \]
This means that one mole of the complex produces one complex ion (Co(NH₃)₅Cl)²⁺ and two moles of chloride ions (Cl⁻).

### Step 3: Reaction with AgNO₃
When excess AgNO₃ is added, it reacts with the chloride ions to form silver chloride (AgCl) precipitate:
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \]
Since we have two moles of Cl⁻ from the dissociation of the complex, they will react with two moles of AgNO₃.

### Step 4: Calculate Moles of AgCl
From the reaction, we see that:
- 1 mole of Cl⁻ produces 1 mole of AgCl.
- Therefore, 2 moles of Cl⁻ will produce 2 moles of AgCl.

### Conclusion
Thus, when Co(NH₃)₅Cl₃ is treated with excess AgNO₃, **2 moles of AgCl** would be obtained.

### Final Answer
**2 moles of AgCl**

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