Consider a 5.313 g mixture of FeO and Fe3O4. You react this mixture with an excess of oxygen to form 5.749 g F

Question

Consider a 5.313 g mixture of FeO and Fe3O4. You react this mixture with an excess of oxygen to form 5.749 g Fe2O3. Calculate the percent by mass of FeO in the original mixture.


help please

Answer 1

Lancenigo di Villorba (TV), Italy

YOUR TROUBLE
You give me a solid iron ores mixture, as FeO/Fe3O4 constitued (e.g. respectively Iron Oxide and Iron TetrOxide or Ferric Ferrite).
You give me weight determination.
This amount is realted to oxides containt as follows :
5.313 = x * 72 + y * 232
where x and y are mole's number respectives for the former and the second oxides, where 72 and 232 are molar weight respectives for the former and the second oxides.
For strong oxidation process you execute the following reactions :
4 FeO(s) + O2(g) ---> 2 Fe2O3(s)
4 Fe3O4(s) + 2 O2(g) ---> 6 Fe2O3(s)
The resulting weight estimated as 5.749, so you calculate the mole number of iron sesquioxide having 160 as molar weight.
You find
x + y = 5.749 / 160 = 0.0359 mol
so you aritmetically put y = 0.0359 - x in the former relation
5.313 = x * 72 + (0.0359 - x) * 232
and by solving you obtain
x = 0.0188 mol and y = 0.0359 - x = 0.0170 mol
or in weight terms expressed

FeO as 1.373 g
and Fe3O4 as 3.940 g.

I hope this helps you.

I hope this helps you.