Calculate the weight of iron which will be converted into its oxide by the action of 18 grn of steam?

Question

give whole process involved

Answer 1

Lancenigo di Villorba (TV), Italy

I AGREE NOT THE PREVIOUS ANSWER.
MY OWN IS 41.9 g.

HISTORICAL NOTEs

At the beginnings of XX century, a german chemist
(F. Haber) proved that PURE FERRITIC or ALPHA-IRON results as a VERY ACTIVE CATALYST in order to convert HOT NITROGEN/HYDROGEN's GAS-STREAM in AMMONIA's One.
Thus, Haber was seeking for a method able to obtain ALPHA-IRON.
Haber discovered that IRON TETROXIDE may be formed by HOT-STEAM ACTION ON IRON's MASS

3 Fe(s) + 4 H2O(g) ---> FeO.Fe2O3(s) + 4 H2(g)

so the Tetroxide backs into PURE ALPHA-IRON BY MEANS OF HYDROGEN's GAS-STREAM

FeO.Fe2O3(s) + 4 H2(g) ---> 3 Fe(s) + 4 H2O(g)



YOU HAVE TO REFER TO FeO.Fe2O3 INSTEAD FeO OR Fe2O3.


CALCULATIONs
The chemical equation is

3 Fe(s) + 4 H2O(g) ---> FeO.Fe2O3(s) + 4 H2(g)

where you retrieve the Stoichiometric Correspondances

nFe : nH2O = (-3) : (-4)

correspondances interssting the Molar Amount ; in the MASS's TERMs will be

mFe : mH2O = [55.9 * (-3)] : [18.0 * (-4)] = (-167.7) : (-72)

Now, I rearrange the terms and I obtain

mFe = 167.7 * mH2O / 72 = 167.7 * 18.0 / 72.0 = 41.9 g

I hope this could be clear.

Answer 2

There are two oxides of iron, FeO and Fe2O3.

18 grams of steam contains 1 mole of oxygen (monatomic) so if you're forming FeO, the amount of iron that could be converted would be 1 mole, or 55.8 g. If you're forming Fe2O3, the most that can be converted would be 0.667 moles, or 37.2 g.

Answer 3

i agree with both the answers, but i would prefer the second one.