# When the formula equation Fe_3O_4 + Al -> Al_2O_3 + Fe is correctly balanced what is the coefficient of Fe?

May 27, 2018

Well...$F {e}_{3} {O}_{4}$ is a mixed valence oxide...$F {e}_{3} {O}_{4} \equiv F e O \cdot F {e}_{2} {O}_{3}$

$3 F {e}_{3} {O}_{4} + 8 A l \rightarrow 9 F e + 4 A {l}_{2} {O}_{3}$

#### Explanation:

And so the oxides are reduced to iron...

$F e O + 2 {H}^{+} + 2 {e}^{-} \rightarrow F e + {H}_{2} O$ $\left(i\right)$

$F {e}_{2} {O}_{3} + 6 {H}^{+} + 6 {e}^{-} \rightarrow 2 F e + 3 {H}_{2} O$ $\left(i i\right)$

And aluminim metal is oxidized to alumina...

$2 A l + 3 {H}_{2} O \rightarrow A {l}_{2} {O}_{3} + 6 {H}^{+} + 6 {e}^{-}$ $\left(i i i\right)$

And we add $3 \times \left(i\right) + 3 \times \left(i i\right) + 4 \times \left(i i i\right)$

$3 F {e}_{2} {O}_{3} + 24 {H}^{+} + 18 {e}^{-} + 3 F e O + 6 {H}^{+} + 6 {e}^{-} + 8 A l + 12 {H}_{2} O \rightarrow 3 F e + 3 {H}_{2} O + 6 F e + 9 {H}_{2} O + 4 A {l}_{2} {O}_{3} + 24 {H}^{+} + 24 {e}^{-}$

And now we cancel common reagents...

${\underbrace{3 F {e}_{2} {O}_{3} + 3 F e O}}_{3 F {e}_{3} {O}_{4}} + 8 A l \rightarrow 6 F e + 4 A {l}_{2} {O}_{3} + \Delta$

And finally....

$3 F {e}_{3} {O}_{4} + 8 A l \rightarrow 9 F e + 4 A {l}_{2} {O}_{3}$

And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....