How many grams of Cacl2 would be needed to precipitate the Ag+ ions from 31.50 mL of a 0.0150 M AgNO3 solution? The molar mass of Cacl2 is 110.98 g/mol.

Answer 1

2AgNO₃(aq) + CaCl₂(aq) ==> 2AgCl(s) + Ca(NO₃)₂(aq) ... balanced equation

mols AgNO₃ present = 31.50 ml x 1 L / 1000 ml x 0.0150 mol/L = 0.0004725 mols AgNO₃ = mols Ag⁺

mols CaCl₂ needed = 0.0004725 mol AgNO₃ x 1 mol CaCl₂ / 2 mol AgNO₃ = 0.0002363 mols CaCl₂ needed

grams CaCl₂ needed = 0.0002363 mols x 110.98 g/mol = 0.0262 g CaCl₂ (3 sig.figs.)