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
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.