Lead (II) Nitrate and potassium iodide are combined.

Question

His many liters of 0.150 M potassium iodide would be required to react completely with 15.00 grams of lead (II) Nitrate?

Answer 1

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

moles Pb(NO3)2 present = 15.00 g x 1 mole/331g = 0.0453 moles

0.0453 mol x 2 mol KI/mol = .0906 mol KI needed

(0.150 mol/L)(xL). = 0.0906 mol

x = 0.604 L