Question

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

Answer 1

`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`