What mass of precipitate (in g) is formed when 70.0 mL of 0.500 M AlI₃ reacts with excess AgNO₃ in the following chemical reaction? AlI₃(aq) + 3 AgNO₃(aq) → 3 AgI(s) + Al(NO₃)₃(aq)

Answer 1

AlI₃(aq) + 3AgNO₃(aq) ==> 3AgI(s) + Al(NO₃)₃(aq)

mols AlI₃ = 70.0 ml x 1 L / 1000 ml x 0.500 mol / L = 0.035 mols AlI₃

mass AgI formed = 0.035 mol AlI₃ x 3 mol AgI / mol AlI₃ x 234.8 g AgI / mol = 24.7 g AgI precipitate (3 s.f.)