Answer:
1.20 mol
Step-by-step explanation:
0.10 mol of NaCl contains 0.10 mol of Cl- and 0.10 mol of CaCl2 contains 0.20 mol of Cl-. So, to precipitate all of the Cl- as AgCl(s), the minimum number of moles of AgNO3 must be 0.20 mol plus 0.10 mol, which equals 1.20 mol.