percentage of potassium iodide in solution

Question

Question 1

1. If we mix 2.39g of unknown mixture containing potassium iodide with excess lead(II) nitrate and 1.21g of PbI₂ precipitate forms what is the percentage of potassium iodide in the unknown mixture?      ___Pb(NO₃)₂ + ___KI à___PbI ₂ + ___KNO₃

Answer 1

2KI(aq) + Pb(NO₃)(aq) ==> PbI₂(s) + 2KNO₃(aq) ... balanced equation

moles of PbI₂ formed = 1.21 g x 1 mol PbI₂/461 g = 0.002625 moles PbI₂

moles KI present in original sample = 0.002625 mol PbI₂ x 2 mol KI/mol PbI₂ = 0.00525 moles KI

mass KI in original sample = 0.00525 mol KI x 166 g/mol = 0.8715 g KI

Percent KI in original sample = 0.8715 g/2.39 g (x100%) = 36.5%

If you want to do this same problem based on the amount of KNO₃ formed, it would look like this:

moles of PbI₂ formed = 1.21 g x 1 mol PbI₂/461 g = 0.002625 moles PbI₂

moles KNO3 = 0.002625 moles PbI2 x 2 mol KNO3/mole PbI2 = 0.00525 moles KNO3

moles KI = 0.00525 mol KNO3 x 2 mol KI/2 mol KNO3 = 0.00525 moles KI

mass KI = 0.00525 mol KI x 166 g/mol = 0.8715 g KI

etc.