Calculate Moles of AgCl Precipitated from Complex Solution

To find the number of moles of $\text{AgCl}$ precipitated, we first determine the chemical formula of the coordination compound tetraaquadichloridochromium(III) chloride.

1. Identification of the Complex Formula:

   • Central metal ion: Chromium(III) ion, $\text{Cr}^{3+}$
   • Ligands in the coordination sphere: Four neutral aqua ligands ($\text{H}_2\text{O}$) and two anionic chlorido ligands ($\text{Cl}^-$)
   • Coordination sphere: $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+$ (Net charge = $+3 - 2 = +1$)
   • Counter ion: One chloride ion ($\text{Cl}^-$) to neutralize the complex cation.

   Therefore, the IUPAC formula of the complex is: $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl}$

2. Ionization in Aqueous Solution: When dissolved in water, only the chloride ion present in the outer ionization sphere dissociates: $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \,(aq) \longrightarrow [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ \,(aq) + \text{Cl}^- \,(aq)$

   This indicates that $1 \text{ mole}$ of the complex produces $1 \text{ mole}$ of ionizable $\text{Cl}^-$ ions.

3. Calculation of Moles of the Complex: Given:

   • Volume of solution, $V = 100 \text{ mL} = 0.100 \text{ L}$
   • Molarity of solution, $M = 0.05 \text{ M}$

   $\text{Moles of complex} = \text{Molarity} \times \text{Volume (L)}$ $\text{Moles of complex} = 0.05 \text{ M} \times 0.100 \text{ L} = 0.005 \text{ mol} = 5 \times 10^{-3} \text{ mol}$

4. Calculation of Moles of $\text{AgCl}$ Precipitated: Reaction of ionizable $\text{Cl}^-$ ions with excess $\text{AgNO}_3$: $\text{Ag}^+ \,(aq) + \text{Cl}^- \,(aq) \longrightarrow \text{AgCl} \,(s)$

   $\text{Moles of AgCl precipitated} = \text{Moles of ionizable Cl}^- \text{ ions} = 5 \times 10^{-3} \text{ mol}$

Thus, the number of moles of $\text{AgCl}$ precipitated is $5 \times 10^{-3}$.

Answer: 5