Question

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?

Answer 1

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) = $C•V =$ 0.1(M)•0.005(L) = 0.0005 mol

$\nu(KI) = C•V$ = 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) = $1/2 • \nu(KI)$ = 1/2 • 0.0002(mol) = 0.0001 mol

Then, the amount of the precipitate is equal to:

m(PbI₂) = $\nu•M$ = 0.0001(mol)•461(g/mol) = 0.05 g

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