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?

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?