m ( KI ) = (2%*500)/100 = 10g
2KI + Pb(NO3)2 = PbI2 + 2 KNO3
v( KI) = 10g/166g/mol = 0.06 mol
v( PbI2 ) = 0,06/2 =0.03 mol
m ( KI ) = (2%*500)/100 = 10g
2KI + Pb(NO3)2 = PbI2 + 2 KNO3
v( KI) = 10g/166g/mol = 0.06 mol
v( PbI2 ) = 0,06/2 =0.03 mol
m(PbI2) = 0,03*461 g/mol = 13,83g