When two moles of `[Co(NH_3)_5Cl]Cl_2` is treated with excess silver nitrate solution, the number of moles of silver chloride formed is

Options

A. 3
B. 6
C. 4
D. 2

Answer

To solve the problem, we need to analyze the reaction between the complex compound `[Co(NH_3)_5Cl]Cl_2` and silver nitrate (`AgNO_3`).

### Step 1: Determine the composition of the complex
The complex `[Co(NH_3)_5Cl]Cl_2` consists of:
- 1 cobalt ion (Co)
- 5 ammonia ligands (NH₃)
- 1 chloride ligand (Cl) coordinated to cobalt
- 2 chloride ions (Cl⁻) in the outer sphere

### Step 2: Identify the number of chloride ions
When the complex dissociates in solution, it releases:
- 1 chloride ion from the coordination sphere (the one directly bonded to cobalt)
- 2 chloride ions from the outer sphere

Thus, for every mole of the complex, there are a total of 3 moles of chloride ions produced (1 from the complex and 2 from the outer sphere).

### Step 3: Calculate the total number of moles of chloride ions from 2 moles of the complex
If we have 2 moles of `[Co(NH_3)_5Cl]Cl_2`, the total number of moles of chloride ions released will be:
\[ \text{Total chloride ions} = 2 \text{ moles of complex} \times 3 \text{ moles of Cl} = 6 \text{ moles of Cl} \]

### Step 4: Determine the reaction with silver nitrate
When silver nitrate is added, each chloride ion reacts with silver ions to form silver chloride (AgCl):
\[ \text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl} \]

Since we have 6 moles of chloride ions, the reaction will produce:
\[ \text{Total moles of AgCl} = 6 \text{ moles of Cl} \rightarrow 6 \text{ moles of AgCl} \]

### Final Answer
Therefore, when 2 moles of `[Co(NH_3)_5Cl]Cl_2` are treated with excess silver nitrate, the number of moles of silver chloride formed is **6 moles**.

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