2KI + Pb(N03)2 → 2KN03 + Pbl2

Question

Bobby and Jane dissolve 26.8 g of solid potassium iodide in water, then pour the solution into an excess of concentrated lead (ll) nitrate. Jane carefully filters the resulting yellow precipitate from the solution and dries the solid overnight. The following day, Bobby finds the mass of the precipitate (which he knows to be Pbl₂) to be  34.8 g. What is the percent yield of Bobby and Jane's reaction?

Answer 1

2KI + Pb(NO₃)₂ ==> 2KN0₃ + Pbl₂

Theoretical yield of PbI2:

26.8 g KI x 1 mol KI / 166 g x 1 mol PbI₂ / 2 mol KI x 461 g PbI₂ / mole PbI₂ = 37.2 g PbI₂

Percent yield = actual yield / theoretical yield (x100%) = 34.8 g / 37.2 g (x100%) = 93.5% yield