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 precipitate when an excess of AgNO₃ is added to a solution of dichloridotetraaquachromium (III) chloride.

### Step-by-Step Solution:

1. **Identify the Chemical Reaction**:
The reaction between dichloridotetraaquachromium (III) chloride and silver nitrate can be written as:
\[ \text{CrCl}_2\text{(H}_2\text{O)}_4 + 2\text{AgNO}_3 \rightarrow \text{Cr(H}_2\text{O)}_4\text{Cl}_2 + 2\text{AgCl} \downarrow + 2\text{NO}_3^- \]
Here, AgCl precipitates out of the solution.

2. **Calculate the Moles of Dichloridotetraaquachromium (III) Chloride**:
Given the concentration and volume of the dichloridotetraaquachromium (III) chloride solution:
- Concentration = 0.2 M
- Volume = 100 mL = 0.1 L

Using the formula:
\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of CrCl}_2 = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

3. **Convert Moles to Millimoles**:
Since 1 mole = 1000 millimoles:
\[ \text{Millimoles of CrCl}_2 = 0.02 \, \text{mol} \times 1000 \, \text{mmol/mol} = 20 \, \text{mmol} \]

4. **Determine the Amount of AgCl Precipitated**:
From the balanced equation, we see that 1 mole of dichloridotetraaquachromium (III) chloride reacts with 2 moles of AgNO₃ to produce 2 moles of AgCl. Therefore, the number of moles of AgCl produced will be equal to the number of moles of dichloridotetraaquachromium (III) chloride:
\[ \text{Millimoles of AgCl} = 20 \, \text{mmol} \, \text{(from CrCl}_2\text{)} \]

5. **Final Answer**:
The number of millimoles of AgCl precipitated is **20 mmoles**.