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 determine how many moles of silver chloride (AgCl) are formed when two moles of the coordination compound \([Co(NH_3)_5Cl]Cl_2\) are treated with excess silver nitrate (AgNO₃).

### Step-by-Step Solution:

1. **Identify the Composition of the Coordination Compound**:
The coordination compound \([Co(NH_3)_5Cl]Cl_2\) consists of one cobalt ion (Co) surrounded by five ammonia (NH₃) ligands and one chloride ion (Cl) within the coordination sphere. The two chloride ions outside the coordination sphere are the ones that will react with silver nitrate.

2. **Determine the Number of Chloride Ions**:
Each molecule of \([Co(NH_3)_5Cl]Cl_2\) contains two chloride ions: one inside the coordination sphere and one outside. Since we have two moles of the compound, we can calculate the total number of chloride ions:
\[ \text{Total chloride ions} = 2 \text{ moles} \times 2 \text{ chloride ions/mole} = 4 \text{ chloride ions} \]

3. **Reaction with Silver Nitrate**:
When silver nitrate is added, it reacts with the chloride ions to form silver chloride (AgCl). The reaction can be represented as:
\[ Ag^+ + Cl^- \rightarrow AgCl \]
Since we have 4 chloride ions available, they will react with 4 moles of silver ions to produce 4 moles of silver chloride.

4. **Calculate the Moles of Silver Chloride Formed**:
Therefore, the total moles of silver chloride formed from the reaction will be equal to the number of chloride ions:
\[ \text{Moles of AgCl formed} = 4 \text{ moles} \]

### Final Answer:
The number of moles of silver chloride formed is **4 moles**.