A six co-ordinate complex has the formula `CoCl_(3).5NH_(3).H_(2)O`. Electrical coductance measurements indicate the presence of three ions in one formula unit. How many moles of `AgCl` will be percipated with excess `AgNO_(3)` solution with two mole of complex?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the complex and its dissociation
The complex given is \( CoCl_3 \cdot 5NH_3 \cdot H_2O \). It is a six-coordinate complex, which means the central metal ion (Co) is surrounded by six ligands. In this case, we have 5 ammonia (NH3) molecules and 1 chloride ion (Cl-) as ligands.

### Step 2: Determine the number of ions produced upon dissociation
The electrical conductance measurements indicate that one formula unit of the complex produces three ions. This implies that:
- 1 complex ion (which contains Co and the ligands)
- 2 chloride ions (Cl-)

Thus, the dissociation can be represented as:
\[ CoCl_3 \cdot 5NH_3 \cdot H_2O \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^- \]

### Step 3: Calculate the number of moles of Cl- produced from 2 moles of the complex
From the dissociation, we see that:
- 1 mole of the complex produces 2 moles of Cl-.

Therefore, if we have 2 moles of the complex:
\[ 2 \text{ moles of complex} \rightarrow 2 \times 2 = 4 \text{ moles of Cl-} \]

### Step 4: Determine the reaction with AgNO3
When chloride ions (Cl-) react with silver nitrate (AgNO3), they form silver chloride (AgCl) as a precipitate:
\[ Ag^+ + Cl^- \rightarrow AgCl \]

Since we have 4 moles of Cl-, the reaction will yield:
\[ 4 \text{ moles of Cl-} \rightarrow 4 \text{ moles of AgCl} \]

### Conclusion
Thus, the total amount of AgCl precipitated when 2 moles of the complex react with excess AgNO3 is:
\[ \text{4 moles of AgCl} \]

### Final Answer
**4 moles of AgCl will be precipitated.**

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