An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromimum (III) chloride. The number of moles of AgCl precipitated would be:

The molecular formula of dichlorotetraaquachromimum (III) chloride can be represented as [Cr(H2O)4Cl2]Cl. Thus, it contains one ionizable Cl⁻.

The number of moles of [Cr(H2O)4Cl2]Cl in 100 mL of a 0.01 M solution is:

(0.01 mol/L) * (100 mL) * (1 L/1000 mL) = 0.001 mol

Since there is one ionizable Cl⁻ ion per formula unit, 0.001 moles of Cl⁻ ions are present in the solution.

When excess AgNO3 is added, it reacts with the Cl⁻ ions to form AgCl precipitate according to the following reaction:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

The mole ratio between Cl⁻ and AgCl is 1:1. Therefore, 0.001 moles of Cl⁻ will produce 0.001 moles of AgCl.

Therefore, the number of moles of AgCl precipitated would be 0.001.