Answer:
336.1 g of PbS precipitate
Explanation:
The equation of the reaction is given as;
Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)
Ionically;
Pb^2+(aq) + S^2-(aq) -----> PbS(s)
Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide
Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles
Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate
Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles
Then we determine the limiting reactant. The limiting reactant yields the least amount of product.
Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide
1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide
Hence sodium sulphide is the limiting reactant.
Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide
Molar mass of lead II sulphide= 287.26 g/mol
Mass of lead II sulphide = 1.17 moles × 287.26 g/mol
Mass of lead II sulphide= 336.1 g of PbS precipitate
