Solution Stoichiometry

Question

How many grams of Ag₂CO₃ will precipitate when excess (NH₄)₂CO₃ solution is added to 63.0 mL of 0.763 M AgNO₃ solution?

2AgNO₃(aq) + (NH₄)₂CO₃(aq) Ag₂CO₃(s) + 2NH₄NO₃(aq)

Answer 1

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

From this balanced equation, note that 1 mole Ag₂CO₃(s) is formed from TWO moles of AgNO₃. This mole ratio is important in solving this problem.

moles AgNO₃ present = 63.0 ml x 1 L/1000 ml x 0.763 moles/L = 0.048069 moles AgNO₃

moles Ag₂CO₃(s) that can form = 0.048069 mol AgNO3 x 1 mol Ag₂CO₃(s) / 2 moles AgNO3 = 0.0240 moles

molar mass Ag₂CO₃ = 275.6 g/mol

mass Ag₂CO₃(s) = 0.0240 moles x 275.6 g/mole = 6.62 g Ag₂CO₃