Question

50.0 mL of a 0.250 M AgNO3 solution and 50.0 mL of 0.400 M Na2CO3 solution are

mixed. How much of the silver carbonate precipitate will be formed?

2 AgNO3 (aq) + Na2CO3 (aq) -> Ag2CO3 (s) + 2 NaNO3 (aq)

Answer 1

2Ag Noz-Laq) + Na, coz lage) Ag, CO3(s) + 2NONO, Millimolis of Agnoz - Concentration x Volume = 0.250 M X 50 mL = 12.5 mmoles illimoles of Na₂CO₃ = 0.40 M x 50ML = 20 mmolus. Imole of Na₂CO3 reacts with 2 moles of AgNO3. Lommeles Na₂CO₃ will react with : - 2 moles AgNO3 2ommel Na, Coz 1 mol Na2CO₃ = 40mmeles AgNO3 But there is only 12.5 mnoles of AgNo so, AgNO3 will be the limiting reagent 2 moles of AgNO3 forms 1 mole of Ag, CO₂ 12.5 mmeles will form: Imo Ag, C03 x 12.5 mmel AgNO₃ 2me AgNO3 = 6.25 mme Ag₂CO3 Mass of Ag CO₂ = 6.25 more x 10 ml x 275.7 gonelt Impel = 1.729 Mass of AgCO3 = 1.72g.