Based on the following equation: 2AgNO3 + CaCl2 > 2AgCl + Ca(NO3) When 50 g AgNO3 are combined with 1.45M CaCl2 solution, what volume of the 1.45M CaCl2 solution would be needed to precipitate all of the silver as AgCl?

Answer 1

mols AgNO3 = grams/molar mass = approx 0.3 but you need a better answer than that estimate.
Convert mols AgNO3 to mols CaCl2. That means approx 0.3 x (1 mol CaCl2/2 mols AgNO3) = approx 0.15 mol CaCl2.
Then M CaCl2 = mols CaCl2/L CaCl2. You have M CaCl2 and mols CaCl2, solve for L CaCl2. Covert to mL if needed.