Question Detail
Question
If 257.8 mL of a 0.0468 M solution of lead nitrate is added to 156.00 mL of a 0.095 M solution of sodium sulphide, what is the mass of solid lead sulphide formed? [ Pb = 207, Na = 23, N=14; O=16; S=32] Remember to balance the equation.
Correct Answer
Option B is the correct answer.
Detailed Explanation
Balanced equation: Pb(NO₃)₂(aq) + Na₂S(aq) → PbS(s) + 2NaNO₃(aq). Moles Pb(NO₃)₂ = (0.2578 L)(0.0468 mol/L) = 0.01206 mol. Moles Na₂S = (0.156 L)(0.095 mol/L) = 0.01482 mol. Pb(NO₃)₂ is the limiting reactant. Moles PbS = 0.01206 mol. Mass PbS = 0.01206 × (207 + 32) = 0.01206 × 239 = 2.882 g ≈ 2.88 g.
Hint
Find the limiting reactant first by calculating moles of each.