Calculate the mass of PbCl2 formed when an excess of 0.100 M solution of NaCl is added to 0.4L of 0.7 M Pb(NO3)2.

Question

Calculate the mass of PbCl2 formed when an excess of 0.100 M solution of NaCl is added to 0.4L of 0.7 M Pb(NO3)2. 
Pb(NO3)2 + 2 NaCl --> PbCl2 + 2 NaNO3

DrBob222 · Answer

mols Pb(NO3)2 = M x L = approx 0.28 mols PbCl2 formed = mols Pb(NO3)2 (see the 1:1 ratio for Pb(NO3)2/PbCl2 Then g PbCl2 = mols PbCl2 x molar mass PbCl2.