Question

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate would you expect to form?

Answer 1

336.1 g of PbS precipitate

Step-by-step 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