How many grams of NaCl would be needed to precipitate the Ag⁺ ions from 2.350 × 10² mL of a 0.0115 M AgNO₃ solution? The molar mass of NaCl is 58.44 g/mol.

Answer 1

AgNO₃(aq) + NaCl(aq) ==> AgCl(s) + NaNO₃(aq) ... balanced equation

mols AgNO₃ = 2.350x10² ml x 1 L / 1000 ml x 0.0115 mol / L = 0.00270 mols AgNO₃

mols NaCl needed = 0.00270 mols AgNO₃ x 1 mol NaCl / mol AgNO₃ = 0.00270 mols NaCl

grams NaCl needed = 0.00270 mols NaCl x 58.44 g NaCl / mol = 0.158 grams NaCl (3 sig. figs.)