Question

what mass of lead (ii) trioxonitrate (v), Pb (NO5)2, would be required to yield 9g of lead (ii) chloride, PbCl, on the addition of excess sodium chloride solution, NaCl? (Pb=207, N=4, O=16, Na=23, Cl=35.5)
Pb(NO3)2 (aq) + 2NaCl(aq)
1 mole 2 moles
331g

PbCl2 (s) + 2NaNO3(aq)
1 mole 2 moles
278g

Answer 1

Who told you that lead (ii) trioxonitrate (v) was the name for Pb(NO3)2? I is not. Please give me a reference. If that is published what is the name of the book? If a teacher what's his/her name and institution. That is NOT a correct name. A correct name is lead(II) nitrate. And note that lead chloride is PbCl2. The equation is not correct either.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3
mols PbCl2 = grams/molar mass = 9 g/278 = 0.0324
From the equation you know that 1 mol PbCl2 is formed from 1 mol Pb(NO3)2; therefore, you will need to start with 0.0324 mols of Pb(NO3)2. Convert that to grams by grams = mols Pb(NO3)2 x molar mass Pb(NO3)2.
Post your work if you get stuck.