Răspuns:
100kg minereu.......69,6kg oxid de fier
1000kg(1 t)..................x
x=696kg oxid
miu,oxid= 232g/mol=232kg/kmol
niu,oxid=696/232kmol=3kmol oxid
1kmol................3kmol
Fe3O4====> 3Fe + 2O2
3kmol...............x
x=niu=9kmol FePUR
m,Fe=9x56kg=504kg FePUR daca randamentul este 100%
la randament 93% se obtine mai putin
m,Fe=504x93/100kg=468,72kg
100KG Fe tehnic........(100-4)kg Fe pur
x..............................................468,72
x=488,25kg Fe tehnic
Explicație: