Precipitate Mass and Heat from a Double-Displacement Reaction
A student mixes 50.0 mL of 0.100 mol/L lead(II) nitrate solution with 50.0 mL of 0.200 mol/L potassium iodide solution in an insulated cup. A yellow precipitate forms. The mixture is stirred; the final volume is 100.0 mL. The initial temperature of both solutions is 20.0 °C and the highest temperature reached is 23.4 °C. Assume the mixed solution has density 1.00 g/mL and specific heat capacity 4.2 J/(g·°C), and that no heat is lost to the cup or surroundings. The precipitate is PbI₂; the other product, KNO₃, remains dissolved. Molar masses: Pb = 207, I = 127, N = 14, O = 16, K = 39.
1. Determine which reactant is limiting and calculate the maximum mass of PbI₂ precipitate.
Read the worked answer
The reaction is:
Pb(NO₃)₂ + 2KI → PbI₂(s) + 2KNO₃
Moles of Pb²⁺ = 0.0500 L × 0.100 mol/L = 0.00500 mol.
Moles of I⁻ = 0.0500 L × 0.200 mol/L = 0.0100 mol.
The reaction needs 2 mol I⁻ per 1 mol Pb²⁺. For 0.00500 mol Pb²⁺, exactly 0.0100 mol I⁻ is needed. The reactants are present in the exact stoichiometric ratio, so neither is left over in excess; either can be used to calculate the product.
Moles PbI₂ = 0.00500 mol.
Molar mass PbI₂ = 207 + 2(127) = 461 g/mol.
Mass = 0.00500 × 461 = 2.305 g ≈ 2.31 g.
2. Calculate the heat released and the heat released per mole of PbI₂ formed. Is the reaction exothermic?
Read the worked answer
The temperature rise is 23.4 − 20.0 = 3.4 °C. The solution mass is about 100.0 g.
q = mcΔT = 100.0 × 4.2 × 3.4 = 1428 J ≈ 1.43 kJ.
This heat was released by the reaction, so the reaction is exothermic. Per mole of PbI₂:
1428 J / 0.00500 mol = 285,600 J/mol ≈ 286 kJ/mol.
If expressed as an enthalpy change, ΔH ≈ −286 kJ/mol.
3. Suppose the student used 50.0 mL of 0.100 mol/L Pb(NO₃)₂ but only 25.0 mL of 0.200 mol/L KI. Predict how the precipitate mass and the temperature rise change. Support with calculations.
Read the worked answer
New moles of I⁻ = 0.0250 L × 0.200 mol/L = 0.00500 mol.
Moles of Pb²⁺ are still 0.00500 mol, but 0.00500 mol Pb²⁺ would need 0.0100 mol I⁻. Now I⁻ is limiting.
Moles PbI₂ = 0.00500 / 2 = 0.00250 mol.
Mass = 0.00250 × 461 ≈ 1.15 g, about half the previous mass.
Heat released ≈ 0.00250 × 285,600 ≈ 714 J. The total solution mass is now about 75 g, so:
ΔT ≈ 714 / (75 × 4.2) ≈ 2.3 °C.
The temperature rise is smaller than 3.4 °C because less reaction occurs and the same heat is spread through a smaller but still substantial mass.