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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#

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#

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Answer 2

Owen Ambrose

The balanced chemical equation for the reaction ( \text{Fe} + \text{H}_2\text{O} = \text{Fe}_3\text{O}_4 + \text{H}_2 ) is ( \text{3Fe} + \text{4H}_2\text{O} = \text{Fe}_3\text{O}_4 + \text{4H}_2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.