You should proof read your questions before posting. I assume you are treating the nitrate with NaCl to yield PbCl2 but I need a crystal ball to guess at that.
Pb(NO3)2 + 2NaCl --> PbCl2 + 2NaNO3
mols PbCl2 = 9.4/molar mass PbCl2
mol PbCl2 = 1mol Pb(NO3)2
Now convert mols Pb(NO3)2 to grams; i.e.,
g = mols x molar mass
Post your work if you get tuck.
What mass of lead pb(No3) would be required ti yield 9g of lead pb(No3)2 would be required to yeild 9g of lead pb cl2,on the addition of excess Nacl
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