The no.of electrons involved in the change,
`Fe_(3)O_(4)rarrFe_(2)O_(3)`:

Fe_(3)O_(4) is:

The equivalent weight of a species if acts as oxidant or reductant should be derived by : Eq. weight of oxidant or reductant = ("Mol. wt. of oxidant or reductant")/{("Number of electrons lost or gained by one"),("moleculae of oxidant or reductant"):} During chemical reactions, equal equivalents of one species react with same number of equivalents of other species giving same number of equivalent of products. However this is not true for reactants if they react in terms of moles. Also Molarity can be converted to normality by multiplying the molarity with valence factor or 'n' factor. Equivalent weight of Fe_(2)O_(3) in terms of its mol. weight in the change Fe_(3)O_(4)rarrFe_(2)O_(3) is

Calculate the number of electrons lost in the following change : Fe + H_(2)O rarr Fe_(3)O_(4)+H_2

The number of electrons lost in the following change is Fe+H_(2)OrarrFe_(3)O_(4)+H_(2)

The number of electrons lost in the following change is Fe+H_(2)OrarrFe_(3)O_(4)+H_(2)

Calculate the n-factor of reactants in the given chemical changes? (d) FeSO_(4)rarr Fe_(2)O_(3)

Calculate the n-factor of reactants in the given chemical changes? (d) FeSO_(4)rarr Fe_(2)O_(3)

The number of electrons lost or gained during the change, Fe + H_2O rarrFe_3O_4 +H_2