How do i find the grams?

Question

Pb(NO₃)₂ (aq) + 2KI (aq) à 2KNO₃ (aq) + PbI₂ (s)

How many grams of lead (II) iodide will precipitate when 211.8 mL of 1.97 M potassium iodide solution is reacted with an excess solution of lead (II) nitrate?

Answer 1

The governing equation for this chemical reaction is the one given:

Pb(NO₃)₂ (aq) + 2KI (aq) → 2KNO₃ (aq) + PbI₂ (s)

This tells us that for every mole of lead iodide (PbI₂) produced, it requires 2 moles of potassium iodide (KI).

A 1.97 M solution of potassium iodide (KI) means the solution contains 1.97 moles per liter of KI and since you have 211.8 mL or 0.2118 liters then the number of moles you have is (1.97)(0.2118) = 0.4172 (moles of KI).

Since it requires 2 moles of KI to produce 1 mole of PbI₂, then there will be 0.4172/2 or 0.2086 moles of PbI₂ produced.

To find the number of grams that is, you must use the molar mass. To calculate the molar mass of PbI₂ go to the periodic table and look up lead (Pb) and iodine (I). PB = 207.2 g/mole and I = 126.9 g/mole but there are 2 atoms of iodine in PbI₂ so the molar mass of PbI₂ = 207.2 + 126.9(2) = 461 g/mole.

Since we have 0.2086 moles of PbI₂ produced and the molar mass of PbI₂ is 461 g/mole then (0.2086)(461) = 96.2 g of PbI₂ will be precipitated (assuming no losses)