How many moles of potassium iodide, KI, are required to precipitate all of the lead (II)ion from 25.0 mL of a 1.6 M Pb(NO₃)₂ solution? (First, write a balanced equation for the reaction.) A)0.040 mole B)0.080 mole C)0.020 mole D)64 mol E)0.64 mol

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3

From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2. Therefore, to precipitate all of the Pb2+ ions from 25.0 mL of 1.6 M Pb(NO3)2 solution, we need:

moles of Pb(NO3)2 = 1.6 M × 0.0250 L = 0.040 mol

moles of KI = 2 × moles of Pb(NO3)2 = 2 × 0.040 mol = 0.080 mol

Therefore, the correct answer is B) 0.080 mole.

How Many Moles of Potassium Iodide, KI, Are Required to Precipitate