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. 0.001
B. 0.002
C. 0.003
D. 0.01

Explanation

To solve the problem, we need to determine the number of moles of silver chloride (AgCl) precipitated when an excess of AgNO₃ is added to a solution of dichlorotetraaquachromium(III) chloride.

### Step-by-Step Solution:

1. **Identify the Compound**: The compound given is dichlorotetraaquachromium(III) chloride. Its formula can be represented as [Cr(H₂O)₄Cl₂]Cl.

2. **Determine the Ionization**: When dichlorotetraaquachromium(III) chloride ionizes in solution, it produces chromium ions and chloride ions:
\[ [Cr(H₂O)₄Cl₂]Cl \rightarrow [Cr(H₂O)₄Cl₂]^{+} + Cl^{-} \]
This means for every 1 mole of the dichlorotetraaquachromium(III) chloride, we get 1 mole of chromium ion and 1 mole of chloride ion.

3. **Calculate the Initial Moles of the Compound**: We have a 0.01 M solution of dichlorotetraaquachromium(III) chloride in a volume of 100 mL. To find the number of moles:
\[ \text{Moles of dichlorotetraaquachromium(III) chloride} = \text{Molarity} \times \text{Volume (L)} \]
\[ = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{moles} \]

4. **Determine the Moles of Chloride Ions Produced**: From the ionization, we know that each mole of dichlorotetraaquachromium(III) chloride produces 1 mole of chloride ions. Therefore, the moles of chloride ions produced will also be 0.001 moles.

5. **Reaction with AgNO₃**: When AgNO₃ is added, it reacts with the chloride ions to form AgCl:
\[ Ag^{+} + Cl^{-} \rightarrow AgCl \]
Since we have an excess of AgNO₃, all the chloride ions will react to form AgCl.

6. **Calculate the Moles of AgCl Precipitated**: Since 1 mole of chloride ions reacts with 1 mole of AgNO₃ to produce 1 mole of AgCl, the number of moles of AgCl precipitated will be equal to the number of moles of chloride ions:
\[ \text{Moles of AgCl} = \text{Moles of Cl}^{-} = 0.001 \, \text{moles} \]

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