Question

When a solution of ammonium chloride, NH4Cl, is added to a solution of lead (II) nitrate , Pb(NO3)2, a white precipitate, PbCl2, forms. Which of the following is the total ionic equation for this reaction?

A) 2 NH4Cl(aq)+Pb(NO3)2(aq)-->PbCl2(s)+2 NH4NO3 (aq)

B) 2 NH4+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)-->PbCl2(s)+2NH4+(aq)+2NO3-(aq)

C) Pb2+(aq)+2Cl-(aq)-->PbCl2(s)

D) Pb2+(aq)+Cl2-(aq)-->PbCl2(s)

Answer 1

2 NH4Cl(aq)+Pb(NO3)2(aq)-->PbCl2(s)+2 NH4NO3 (aq)   molecular equation

2 NH4+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)-->PbCl2(s)+2NH4+(aq)+2NO3-(aq)    total ionic equation

Removal of spectator ions to get net ionic equation

Pb2+(aq)+2Cl-(aq)-->PbCl2(s) Net ionic equation

B) 2 NH4+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)-->PbCl2(s)+2NH4+(aq)+2NO3-(aq) >>>>>answer