Respuesta :

Hagrid

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

Otras preguntas