An excess of `AgNO_(3)` solution is added to 100 mL of a 0.2 M dichloridotetraaquachromium (III) chloride. The numberof millimoles of `AgCl` precipitated would be ----.

Answer

To solve the problem, we need to determine the number of millimoles of AgCl that will precipitate when an excess of AgNO₃ is added to a solution of dichloridotetraaquachromium (III) chloride. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Complex
We have dichloridotetraaquachromium (III) chloride, which can be represented as [Cr(H₂O)₄Cl₂]Cl. This indicates that the chromium (Cr) is in a +3 oxidation state, surrounded by four water molecules (H₂O) and two chloride ions (Cl⁻).

### Step 2: Determine the Charge of the Complex
The overall charge of the complex can be calculated as follows:
- The oxidation state of Cr in the complex is +3.
- Each Cl⁻ contributes -1, and there are two Cl⁻ ions.
- The water molecules (H₂O) are neutral.

Thus, the overall charge of the complex is:
\[ \text{Charge} = +3 + (-1 \times 2) + 0 = +1 \]

### Step 3: Identify the Counter Ion
Since the complex has a +1 charge, it will have one chloride ion (Cl⁻) as a counter ion to balance the charge. Therefore, the formula can be simplified to [Cr(H₂O)₄Cl₂]⁺.

### Step 4: Reaction with AgNO₃
When AgNO₃ is added, the chloride ions will react with Ag⁺ to form AgCl precipitate. The reaction can be represented as:
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \]

### Step 5: Calculate the Number of Moles of the Complex
We need to find the number of moles of the dichloridotetraaquachromium (III) chloride complex in the solution. Given:
- Volume = 100 mL = 0.1 L
- Concentration = 0.2 M

Using the formula:
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Number of moles} = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

### Step 6: Convert Moles to Millimoles
To convert moles to millimoles:
\[ \text{Millimoles} = \text{Number of moles} \times 1000 \]
\[ \text{Millimoles} = 0.02 \, \text{mol} \times 1000 = 20 \, \text{mmol} \]

### Step 7: Conclusion
Since each mole of the complex will produce one mole of AgCl, the number of millimoles of AgCl precipitated will also be 20 mmol.

### Final Answer
The number of millimoles of AgCl precipitated would be **20 mmol**.

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