What mass of lead (v) trioxonitrate (v) pb (Na3)2 would be required to yield a gram of lead (v) chloride pbcl2, on the addition of excess sodium chloride solution Nacl? (Pb = 207, N= 14, o= 16, Na= 23, cl= 35.5)
You have the wrong name and the wrong formula for Pb(NO3)2. An acceptable name by the International union of Pure and Applied Chemistry (IUPAC) is lead(ii) nitrate. PbCl2 is lead (ii) nitrate.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3.
moles PbCl2 needed = grams/molar mass = 1/278 = 0.0036
mols Pb(NO3)2 = 0.0036 mols PbCl2 x (1 mol Pb(NO3)2/1 mol PbCl2) = 0.0036 x 1/1 = 0.0036
grams Pb(NO3)2 = mols Pb(NO3)2 x molar Pb(NO3)2 = ? grams