• Asignatura: Química
  • Autor: Ana033798
  • hace 9 años

Conposición porcentual de:
H3 PO4
FE3 (PO4)2
Mg (OH)2
SO2
Por faa

Respuestas

Respuesta dada por: snorye
6
H3 PO4
Mm H3PO4
H: 3 x 1 = 3 g/mol
P:  1 x 31 = 31 g/mol
O: 4 x 16 = 64 g/mol
````````````````````````````
       Mm = 98 g/mol

H: 
98 g ---- 100 %
  3 g ------ x
     x = 3.06 %
P:
98 g ----  100 %
31 g ----  x
     x = 31.63 %
O:
98 g ---- 100 %
64 g ----  x
  x = 65.30 %


Fe3 (PO4)2
Mm
Fe: 3 x 56 = 168 g/mol
 P: 2 X 31 = 62 g/mol
 O: 8 x 16 = 128 g/mol
````````````````````````````````
           Mm = 358 g/mol

Fe:
358 g ---- 100 %
 168 g ----  x
     x = 46.92 %
P:
358 g ---- 100 %
 62 g ----  x
     x = 17.31 %

O:
358 g ---- 100 %
 128 g ----  x
     x =35.75 %



Mg (OH)2
Mm
Mg = 1 x 24 = 24 g/mol
O: 2 x 16 =32 g/mol
H: 2 x  1 = 2 g/mol
````````````````````````````
           Mm = 58 g/mol

Mg
58 g --- 100 %
24 g ---- x
 x = 41.38 %
O:

58 g --- 100 %
32 g ---- x
 x =
55.17 %
H:

58 g --- 100 %
  2 g ---- x
 x = 3.45 %

SO2

Mm
S: 1 x 32 = 32 g/mol
O: 2 x 16 = 32 g/mol
``````````````````````````````
             Mm = 64 g/mol
S:
64 g ----- 100 %
32 g -----  x
  x = 50 %
O:
 64 g ----- 100 %
32 g -----  x
  x = 50 %






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