Question

¿Cual es la composición centesimal del permanganato de potasio (KMnO4)?

Answer 1

calcular Mm KMnO4
  K: 1 x 39 = 39 g/molMn: 1 x 54.93 = 55 g/mol  O: 4 x 16 = 64 g/mol`````````````````````````````````````````         Mm = 158 g/mol
Potasio
      158 g   ------  100 %        39 g   ------      x
           x = 24.68 % K
Manganeso
       158 g   ------  100 %        55 g   ------      x
           x = 34.81 % Mn

Oxigeno
        158 g   ------  100 %          64 g   ------      x
           x = 40.50 % O