How many grams of Ag2CO3 will precipitate when excess (NH4)2CO3 solution is added to 54.0 mL of 0.667 M AgNO3 solution? g

Answer 1

First, write the correctly balanced equation for the reaction taking place...

(NH₄)₂CO₃(aq) + 2AgNO₃(aq) ==> Ag₂CO₃(s) + 2NH₄NO₃(aq) ... balanced equation

Since (NH₄)₂CO₃ is present in excess, the moles of AgNO₃ will determine the theoretical yield of Ag₂CO₃ since it will be the limiting reactant.

moles AgNO₃ present = 54.0 ml x 1 L / 1000 ml x 0.667 mol / L = 0.03602 mols AgNO₃

Grams of Ag₂CO₃ formed (use the mol ratio of the balanced equation and molar mass Ag₂CO₃ to find this value)

0.03602 mols AgNO₃ x 1 mol Ag₂CO₃ / 2 mols AgNO₃ x 275.7 g Ag2CO3 / mol = 4.97 g Ag₂CO₃ (3 s.f.)