Calculate the mass of the precipitate formed by the interaction of a solution containing lead (II) nitrate weighing 66.2 g

Calculate the mass of the precipitate formed by the interaction of a solution containing lead (II) nitrate weighing 66.2 g with an excess of sodium sulfide solution.

Pb (NO3) 2 + Na2S = PbS + 2NaNO3
Let’s find the amount of lead nitrate substance:
n (Pb (NO3) 2) = m (Pb (NO3) 2) / M (Pb (NO3) 2) = 66.2 / 331 = 0.2 mol;
According to the stoichiometry of the reaction:
n (PbS) = n (Pb (NO3) 2) = 0.2 mol;
m (PbS) = n (PbS) * M (PbS) = 0.2 * 239 = 47.8 g.
Answer: m (PbS) = 47.8 g.