When aqueous lead (II) nitrate is added to aqueous potassium iodide, the brillia

Question

When aqueous lead (II) nitrate is added to aqueous potassium iodide, the brilliant yellow solid, lead (II) iodide forms (as well as aqueous potassium nitrate).

a) Write out the balanced chemical equation for this precipitation reaction.

b) Would you predict an insoluble precipitate, considering the two reactants added together? Why or why not?

c) If 100.0 mL of a 0.10 M solution of each reactant is added together, how many grams of lead (II) iodide could be produced?
(Hints: This is a limiting reactant problem. Use the volume and molarity provided to calculate the given moles of each reactant. Then, determine how many moles of the product, lead (II) iodide could be produced from the given quantity of each reactant.)

Explanation / Answer

a)

balance

Pb(NO3) + KI =PbI2 + KNO3

balance

Pb(NO3) + 2KI =PbI2 + 2KNO3

b)

PbI2 is not soluble, since lead halides are NOT soluble

c)

mmol of Pb(NO3)2 = MV = 0.1*100 = 10 mmol

mmol of KI = MV = 0.1*10 = 10 mmol

note that ratio is 1:2 so

10 mmol of KI react with 10/2 = 5 mmol of Pb(NO3)2

mmol of PbI2 possible to form = 5*10^-3

mass = mmol*M;W = (5*10^-3)*461.00894 = 2.305 g of PbI(s) can form

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.