What is the minimum volume of 0.20 M lead(II) nitrate solution required to react with all of the iodide ion in 16.6 grams of potassium iodide dissolved in a solution?

Question

A) 1.0 L

B) 0.75 L

C) 0.10 L

D) 0.50 L

E) 0.25 L

Please explain and show work!!

Answer 1

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

16.6 g KI x (1 mol KI / 166.00 g KI) = 0.100 mol KI

2 mol KI reacts with 1 mol of Pb(NO₃)₂

0.100 mol KI x (1 mol Pb(NO₃)₂ / 2 mol KI) = 0.0500 mol Pb(NO₃)₂

Molarity = mol / L

L = mol / Molarity

L = 0.0500 x (1 L / 0.20 mol) = 0.25 L (answer choice E)