Question

What is the minimum mass of potassium carbonate that must be added to 100.0 mL of a 0.235 M solution of silver(I) nitrate in order to form a precipitate?

Answer 1

The balanced equation is : 2 AgNO₃ (aq)+ K₂CO₃ (aq) Ag₂CO₃ (s) + 2 KNO₃ (aq)

Number of moles of AgNO₃ is , n = Molarity x voluime in L

                                                 = 0.235 M x 0.10 L

                                                 = 0.0235 moles

According to the balanced equation,

2 moles of AgNO₃ reacts with 1 mole of K₂CO₃

0.0235 moles of AgNO₃ reacts with M mole of K₂CO₃

M = ( 0.0235x1) / 2

    = 0.01175 moles

Molar mass of K₂CO₃ is = (2xAt.masss of K) + At.mass of C + (3xAt.mass of O)

                                    = (2x39)+12+(3x16)

                                    = 138 g/mol

We know that number of moles , n = mass / molar mass

So mass of K₂CO₃ neede , m = number of moles x molar mass

                                             = 0.01175 mol x 138 (g/mol)

                                              = 1.621 g