Question

composición centesimal fe2(c2o4)3

Answer 1

1. Calcular Mm
Fe: 2 x 56 = 112 g/mol  C: 6 x 12 = 72 g/mol
  O: 12 x 16 = 192 g/mol````````````````````````````````````           Mm = 376 g/mol
Hierro
      376 g ---- 100 %      112 g ----     x
           x = 29.78 % Fe

Carbono
       376 g ---- 100 %         72 g ----     x
           x = 19.14 % C
Oxigeno
      376 g ---- 100 %      192 g ----     x
           x = 51.06 % O