Answer to Question #110915 in Inorganic Chemistry for David

Question #110915
5 ml of 0.1M pb(NO3)2 is mixed with 10 ml of 0.02 M KI.The amount of pbI2 precipitated is about?
1
Expert's answer
2020-04-20T15:07:33-0400

Pb(NO3) + 2KI PbI2 + 2KNO3

  • For the beginning, we should calculate the amounts of reagents' substances in order to find out which one is in excess:

(Pb(NO3)2) = 0.1(M)•0.005(L) = 0.0005 mol

= 0.02(M)•0.01(L) = 0.0002 mol;

(Pb(NO3)2) > (KI) lead nitrate is in excess, then the amount of substance of PbI2 will be calculated by this formula:

(PbI2) = = 1/2 • 0.0002(mol) = 0.0001 mol

Then, the amount of the precipitate is equal to:

m(PbI2) = = 0.0001(mol)•461(g/mol) = 0.05 g

Answer: (PbI2) = 0.0001 mol; m(PbI2) = 0.05 g

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS