Yes, a white precipitate of lead(II) chloride will form.
The overall equation is:
#Pb(NO_3)_(2(aq)) + 2NaCl_text((aq)rarr PbCl_(2(s)) +2NaNO_(3(aq))#
Lead(II) chloride is insoluble so the ionic equation is:
#Pb_((aq))^(2+)+2Cl_((aq))^(-)rarrPbCl_(2(s))#
The other ions take no part in the reaction so are spectators.
The number of moles of #Pb_((aq))^(2+)# = 0.15 x 0.1 = 0.015
The number of moles of #Cl_((aq))^-# ions added = 0.1 x 0.2 = 0.02
0.015 moles of #Pb^(2+)# require 0.015 x 2 = 0.03 moles #Cl^-# ions.
So there is insufficient #Cl^-# ion to react with all the #Pb^(2+)# so there will be an XS of #Pb^(2+)# and 0.02/2 = 0.01 moles of #PbCl_2# will form.
