Chapter 4 - Section 4.6 - Aqueous Reactions and Chemical Analysis - Checkpoint - Page 171: 4.6.1

(a) 12.3 g

n (mol) of NaCl = mass (g)/Mw (g/mol) = 5/58.5 = 0.0855 mol When 0.0855 mol NaCl treated with enough amount of AgNO₃, it yields 0.0855 mol of AgCl as well according to the following reaction: NaCl + AgNO₃ → AgCl (s) + NaNO₃ → mass of AgCl = n (mol)×Mw (g/mol) = 0.0855 * 143.5 = 12.26 g ≈ 12.3 g