Answer
(a) 12.3 g
Work Step by Step
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