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 mass of solid lead sulphide will be formed? [Pb = 207, Na = 23, N = 14, O = 16, S = 32]. Remember to balance the equation: Pb(NO₃)₂(aq) + Na₂S(aq) → PbS(s) + NaNO₃(aq)
Correct Answer
Option B is the correct answer.
Detailed Explanation
The balanced equation is Pb(NO₃)₂(aq) + Na₂S(aq) → PbS(s) + 2NaNO₃(aq). Moles of Pb(NO₃)₂ = 0.0468 mol/L × 0.2578 L = 0.01206 mol. Moles of Na₂S = 0.095 mol/L × 0.15600 L = 0.01482 mol. The reaction is 1:1, so Pb(NO₃)₂ is the limiting reagent. Moles of PbS produced = 0.01206 mol. Molar mass PbS = 207 + 32 = 239 g/mol. Mass of PbS = 0.01206 mol × 239 g/mol = 2.88 g.
Hint
Balance the equation, calculate the moles of each reactant, identify the limiting reagent, and convert moles of product to mass.