To find the mass of the precipitate (ps) formed when 120g of Pb(NO3)2 reacts with HCl, we first need to balance the chemical equation:

Pb(NO3)2 + 2HCl = PbCl2 + 2HNO3

From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of HCl to form 1 mole of PbCl2.

1 mole of Pb(NO3)2 has a molar mass of 331.21g.
120g of Pb(NO3)2 is equal to 120g / 331.21 g/mol = 0.3621 moles of Pb(NO3)2.

Since the reaction is in 1:1 ratio, the same amount of moles of Pb(NO3)2 will form PbCl2.
Therefore, the mass of PbCl2 formed is:
0.3621 moles * (207.2g/mol) = 75.12g

Therefore, the mass of the precipitate (ps = PbCl2) formed when 120g of Pb(NO3)2 reacts with HCl is 75.12g.