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

1 Answer
May 27, 2018

Answer:

Well...#Fe_3O_4# is a mixed valence oxide...#Fe_3O_4-=FeO*Fe_2O_3#

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

Explanation:

And so the oxides are reduced to iron...

#FeO+2H^+ +2e^(-)rarrFe + H_2O# #(i)#

#Fe_2O_3 +6H^+ +6e^(-)rarr2Fe+3H_2O# #(ii)#

And aluminim metal is oxidized to alumina...

#2Al +3H_2OrarrAl_2O_3+6H^+ +6e^-# #(iii)#

And we add #3xx(i)+3xx(ii)+4xx(iii)#

#3Fe_2O_3 +24H^+ +18e^(-)+3FeO+6H^+ +6e^(-)+8Al +12H_2Orarr3Fe + 3H_2O+6Fe+9H_2O+4Al_2O_3+24H^+ +24e^-#

And now we cancel common reagents...

#underbrace(3Fe_2O_3 +3FeO)_(3Fe_3O_4) +8Al rarr6Fe +4Al_2O_3+Delta#

And finally....

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_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....