Răspunsuri

2014-05-19T14:11:00+03:00
Calculezi masele molare pt fiecare:
Fe3O4 => 3*56+4*16=232 g/mol
Fe2O3 => 2*56+3*16=160 g/mol
FeO => 56+16=72g/mol
%Fe din Fe3O4 = 3*56/232*100 = 72,414 %
%Fe din Fe2O3 = 2*56/160*100 = 70%
%Fe din FeO = 56/72*100 = 77,77%
Cel mai mare continut de fier il are oxidul de fier II FeO
4 3 4
2014-05-19T17:27:48+03:00
M FeO=A Fe + A O 56+16=72 u.a.m. 1 mol Fe = 72 72g FeO...100% 56...x x = 56*100supra 72= 77.(7)% M Fe2O3 = 2A FE +3A O 112+48=160 u.a.m 1 mol de Fe2O3=160g 160 g FE2O3...100% 112....x X= 112*100supra160= 70 %