Question

11,2KOH
Fe Cl2
Fe(OH)2= ? g,moli

Answer 1

Răspuns: 2KOH + FeCl2 = Fe(OH)2 + 2KCl

M KOH = 39+16+1=56g/mol

MFe(OH)2 = 56+2*16 + 1*2 = 90g/mol

Explicație:

11,2g KOH ________x g Fe(OH)2

2*56g KOH _______90 g Fe(OH)2

x=9g Fe(OH)2

n=m/M = 9/90=0,1 moli Fe(OH)2