Question

What mass of AgCl can be prepared by the reaction of 100.0 ml of 0.20 M aqueous AgNO3 with 100.0 ml of 0.15 M CaCl2? Write the molecular and net ionic equations.

Answer 1

no of moles fo AgNO3 = molarity * volume in L

                                    = 0.2*0.1 = 0.02 moles

no of moles of CaCl2   = molarity * volume in L
                                    = 0.15*0.1    = 0.015 moles

                 2AgNO3(aq) + CaCl2(aq) ----------------> 2AgCl(s) + Ca(NO3)2 (aq)

1 mole of CaCl2 react with 2 moles of AgNO3

0.015 moles of CaCl2 react with = 2*0.015/1 = 0.03 moles of AgNO3 is required

AgNO3 is limiting reacgent

2 moles of AgNO3 react with excess of CaCl2 to gives 2 moles of AgCl

0.03 moles of AgNO3 react with excess of CaCl2 to gives = 2*0.03/2 = 0.03 moles of AgCl

mass of AgCl = no of moles * gram molar mass

                        = 0.03*143.32

                         = 4.3g of AgCl >>>>>answer

2AgNO3(aq) + CaCl2(aq) ----------------> 2AgCl(s) + Ca(NO3)2 (aq) molecular equation

2Ag^+(aq) + 2NO3^-(aq) + Ca^2+(aq) + 2Cl^-(aq) ----------------> 2AgCl(s) + Ca^2+(aq) + 2(NO3)^- (aq)

removal of spectator ions to get net ionic equation

2Ag^+(aq) + 2Cl^-(aq) ----------------> 2AgCl(s)

Ag^+(aq) + Cl^-(aq) ----------------> AgCl(s)     Net ionic equation