What mass precipitate will form of 1.50 l of highly concentrated pb(clo3)2 is mixed with 0.300 l of 0.110 m nai

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

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