Question

determine los porcentajes de hierro en feco3, fe2o3 y fe3o4. b) ¿cuántos kilogramos de hierro se podrían obtener a partir de 2.000 kg de fe2o3?

Answer 1

FeCO3
Fe. 1 x 56 = 56 g/mol
 C: 1 x 12 = 12 g/mol
  O: 3 x 16 = 48 g/mol
``````````````````````````````````
           Mm = 116 g /mol
hierro:
116 g ------- 100%
  56 g ------      x
x = 48.28 %

carbono:
116 g ------- 100%
  12 g ------      x
x = 10.34 %

oxigeno:
116 g ------- 100%
  48 g ------      x
x = 41.38  %


Fe2O3
Fe: 2 x 56 =112 g/mol
 O: 3 x 16 = 48 g/mol
````````````````````````````````
           Mm = 160 g/mol


 Fe3O4
Fe: 3 x 56 =168 g/mol
 O: 4 x 16 = 64 g/mol
````````````````````````````````
           Mm = 232 g/mol

b) Mm Fe3O4 = 232 g/mol
                 Fe = 56 g/mol x 3 = 168 g/mol

 232 g Fe3O4 -----  168 g Fe
 100 g  Fe3O4-----    X

         X = 72.41 g

Answer 2

Respuesta:

FeCO3

Fe. 1 x 56 = 56 g/mol

C: 1 x 12 = 12 g/mol

 O: 3 x 16 = 48 g/mol

``````````````````````````````````

          Mm = 116 g /mol

hierro:

116 g ------- 100%

 56 g ------      x

x = 48.28 %

carbono:

116 g ------- 100%

 12 g ------      x

x = 10.34 %

oxigeno:

116 g ------- 100%

 48 g ------      x

x = 41.38  %

Fe2O3

Fe: 2 x 56 =112 g/mol

O: 3 x 16 = 48 g/mol

````````````````````````````````

          Mm = 160 g/mol

Fe3O4

Fe: 3 x 56 =168 g/mol

O: 4 x 16 = 64 g/mol

````````````````````````````````

          Mm = 232 g/mol

b) Mm Fe3O4 = 232 g/mol

                Fe = 56 g/mol x 3 = 168 g/mol

232 g Fe3O4 -----  168 g Fe

100 g  Fe3O4-----    X

        X = 72.41 g

Explicación: