An excess of `AgNO_(3)` is added to `100mL` of a `0.01M` solution of dichlorotetraaquachromium(III) chloride The number of moles of `AgCI` precipitated would be .

Options

A. `0.003`
B. `0.01`
C. `0.0001`
D. `0.002`

Explanation

To solve the problem, we need to calculate the number of moles of AgCl precipitated when an excess of AgNO₃ is added to a 100 mL solution of 0.01 M dichlorotetraaquachromium(III) chloride.

### Step-by-Step Solution:

1. **Identify the complex and its formula**:
The complex is dichlorotetraaquachromium(III) chloride. The formula can be written as [Cr(H₂O)₄Cl₂]Cl. Here, Cr is in the +3 oxidation state.

2. **Determine the charge of the complex**:
The charge on the chromium ion (Cr) is +3. The water molecules (H₂O) are neutral (0 charge), and the two chloride ions (Cl) each have a -1 charge. Thus, the total charge of the complex is:
\[ +3 + 0 \times 4 - 1 \times 2 = +3 - 2 = +1 \]
Therefore, the overall charge of the complex is +1.

3. **Determine the number of chloride ions needed**:
Since the overall charge of the complex is +1, one chloride ion (Cl⁻) is required to balance the charge. Thus, for every mole of the complex, one mole of AgCl will be produced when AgNO₃ is added.

4. **Calculate the number of moles of the complex**:
The number of moles of the complex can be calculated using the formula:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
Given that the volume is 100 mL (which is 0.1 L) and the molarity is 0.01 M:
\[ \text{Number of moles} = 0.1 \, \text{L} \times 0.01 \, \text{M} = 0.001 \, \text{moles} \]

5. **Calculate the number of moles of AgCl precipitated**:
Since 1 mole of the complex produces 1 mole of AgCl, the number of moles of AgCl precipitated will also be:
\[ \text{Number of moles of AgCl} = 0.001 \, \text{moles} \]

### Final Answer:
The number of moles of AgCl precipitated is **0.001 moles**.