se ia
a mol FeO cu M= 72g/mol
b mol Fe2O3 cu M= 160g/mol
m,amstec= 72a +160b
m,Fe din acest amstec=56a+2x56b
72a+160b,g amestec....contine.....56a+112b ,g Fe
1g.............................................................0,724g
0,724(72a+160b)= 56a+112b-------------------> a=b
daca a=b=1mol
m,amestec= 232g
in 232g amestec.........72g FeO.........160gFe2O3
100g,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,x..........................y
calculeaza x si y -procentele masice !!!!!!
