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# 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
2020-04-20T15:07:33-0400

Pb(NO3) + 2KI "\\to" PbI2 + 2KNO3

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

"\\nu"(Pb(NO3)2) = "C\u2022V =" 0.1(M)•0.005(L) = 0.0005 mol

"\\nu(KI) = C\u2022V" = 0.02(M)•0.01(L) = 0.0002 mol;

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

"\\nu"(PbI2) = "1\/2 \u2022 \\nu(KI)" = 1/2 • 0.0002(mol) = 0.0001 mol

Then, the amount of the precipitate is equal to:

m(PbI2) = "\\nu\u2022M" = 0.0001(mol)•461(g/mol) = 0.05 g

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

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