Question

You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from 13.0 mL of 0.180 M AgNO3 solution?

Answer 1

Solution :-

Balanced reaction equation

KCl(aq) + AgNO3(aq) ----- : AgCl(s) + KNO3(aq)

lets first calculate the moles of the AgNO3 using the molarity and volume

moles = molarity * volume in L

moles of AgNO3 = 0.180 mol per L * 0.0130 L                                (13 ml = 0.013 L)

                          = 0.00234 mol AgNO3

now lets find the moles of KCl needed

mole ratio of the KCl and AgNO3 is 1 :1

therefore moles of the KCl needed are same as moles of the AgNO3

hence moles of KCl = 0.00234 mol

now lets convert the moles of the KCl to its mass

mass = moles * molar mass

molar mass of KCl = 74.55 g per mol

mass of KCl = 0.00234 mol * 74.55 g per mol = 0.174 g KCl

Therefore the mass of KCl needed to precipitate the AgCl is 0.174 g KCl