You heat 3.854 g of a mixture of fe3o4 and feo to form 4.148 g fe2o3 the mass percent of the feo in the mixture is

the mass percent of FeO = 24.21 %

Explanation:

according to the balanced reaction equation:

FeO + Fe3O4 + 1/2 O2---> 2Fe2O3

from the balanced equation we can see that 2 moles of  Fe2O3 formed when 1 mole of FeO react with 1 mole of Fe3O4 and with 1/2 mole of oxygen gas

So, the molar ratio between FeO and Fe2O3 is 1:2

* then, we need to number of moles of Fe2O3 first by using this formula:

number of moles (n) = mass of Fe2O3 / molar mass of Fe2O3

the mass is a given = 4.148 g

the molar mass of Fe2O3 = 159.69 g/mol

by substitution:

n = 4.148/159.69 = 0.02598 moles

now, by using the molar ratio between FeO & Fe2O3 we will get the number of moles of FeO

1 mol FeO > 2 mol Fe2O3

          X   < 0.02598 mol

X = 0.01299 moles

now, we can get the mass of FeO by using this formula:

mass of FeO= number of moles FeO * molar mass of FeO

mass of FeO = 0.01299 moles * 71.844 g/mol

                     = 0.933 g

the mass percent of FeO = mass of FeO / Total mass pf mixture *100

                                         = 0.933 g / 3.854 * 100

                                         = 24.21 %