Consider 5.430g Mixture Of FeO And Fe3O4. You React This Mixture With Excess Oxygen To Form 5.779g Fe2O3. Can You Calculate The Percent By Mass Of FeO In The Original Mixture?

5.430g FeO And Fe3O4 mix + O2(excess)  --> 5.779g 2Fe2O3 + O2 left ?

Tough balancing act but if right you have 5.779g/319.2g/mole=0.0181 mole product. You then have 0.009 mole of FeO or x 71.8g/mole = 0.65g and 0.0181mole Fe3O4 x 231.4g/mole = 4.188g. [0.65g/(4.188g+0.65g)]x100%=
13.4 %.


Q: Can You Calculate The Percent By Mass Of FeO In The Original Mixture?