How do you balance the chemical equation: #Fe+H_2O = Fe_3O_4+H_2#?

1 Answer
Apr 28, 2016

Answer:

#3Fe+4H_2O = Fe_3O_4+4H_2#

Explanation:

The #Fe_3# in #Fe_3O_4# on the right implies that there must be some multiple of #3# #Fe# on the left side.

Similarly the #O_4# on the right side implies that there must be some multiple of #4# #H_2O# on the left side.

#(3m)Fe+(4n)H_2O = (p)Fe_3O_4+(q)H_2#

Since
#color(white)("XXX")(3m)Fe rarr (m)Fe_3O_4# (there's no place else for the #Fe# to go)
and
#color(white)("XXX")(4n)H_2O rarr (n)Fe_3O_4# (there's no place else for the #O# to go)
#rArr m=n#

Further
#color(white)("XXX")(4n)H_2O rarr (q)H_2#
#rArr q=4n#

Therefore we have
#color(white)("XXX")(3n)Fe+(4n)H_2O = (n)Fe_3O_4+(4n)H_2#

Using the simplest version: #n=1# gives
#color(white)("XXX")3Fe+4H_2O = Fe_3O_4+4H_2#