Given:
1.50 L of highly concentrated
Pb(Clo3)2
0.300 L of 0.110 M NaI
Required:
mass precipitate of highly
concentrated Pb(Clo3)2
Solution:
M1V1 = M2V2
M1(1.50 L of highly concentrated
Pb(Clo3)2) = (0.300 L)(0.110 M NaI)
M1 = 0.022M Pb(Clo3)2 = 0.022
mol/L Pb(Clo3)2
Molar mass of Pb(Clo3)2= 374.2
g/mol
Mass of Pb(Clo3)2 = (0.022 mol/L
Pb(Clo3)2)( 1.50 L)( 374.2 g/mol) = 12.35g Pb(Clo3)2
