•Peso atómico Pb=207,2 u.m.a.

•Peso atómico Cl=35,453 u.m.a.

•Peso molecular PbCl2=(207,2+2×35,453) u.m.a.=278,106 u.m.a.

•Número de gramos de cloro en PbCl2=

=[(2×35,453 u.m.a)/(278,106 u.m.a.)]×(5 g)=1,27 g.

•Solución:

•1,27 gramos.

(5,0 g PbCl2) / (278,106 g PbCl2/mol) x (2 mol Cl / 1 mol PbCl2) x

(35,4532 g Cl/mol) = 1,2748 g = 1,3 g Cl