If 70.0 mL of 0.150 M CaCl2 is added to 35.0 mL of 0.100 M AgNO3, what is the mass in g of AgCl precipitate?

Question

If 70.0 mL of 0.150 M CaCl2 is added to 35.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate

Answer 1

Please convert everything to molar mass:
70.0 mL of 0.150 M CaCl2 is 0.0105 Mol of CaCl2.
35.0 mL of 0.100 M AgNO3 is 0.0035 Mol of AgNO3
0.0105 Mol of CaCl2 can react with twice as much AgNO3, since it can provide two Cl- ions from one CaCl2.
Thus, the maximum AgCl can be produced is limited by 0.0035 Mol of AgNO3 to 0.0035 Mol of AgCl, which is (0.0035 Mol)*(143.32g/Mol) = 0.502g of AgCl.