a) 2AL + Fe2O3 = Al2O3 + 2Fe
b) MFe2O3 = 2AFe+3AO=2*56+3*16=160g/mol
mO=3*16=48g
%O=mO/MFe2O3*100 = 48/160*100 = Calculeaza !!
c) MFe=56g/mol
n=m/M=188/56=3,3571 moli Fe
1 mol Fe....................6,022*10^23 atomi
3,3571 moli Fe...............x=Calculeaza!!!