A 1.0 L sample of an aqueous solution contains .1 mol of NaCl and .1 mole of CaCl2. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl(1- charge) as AgCl?(assume that AgCl is insoluble.)

Answer 1

Cl^- in NaCl = 0.1 mol
Cl^- in CaCl2 = 0.1 mol x 2 = 0.2 mol

Total Cl^- = 0.3 mol/1 L = 0.3 M

AgNO3 + Cl^- ==> AgCl + NO3^-

Cl^- = 0.3 mols
So what must AgNO3 be to react with all 0.3 mol Cl^-?