Answer:
The mass of Fe₂O₃ in 0.500 g of mixture is 0.367 g.
Explanation:
First off, we know that 72% of the mass of the mixture is iron. The information also tells us that the remaining 28% of the mass is oxygen.
Now we calculate the total mass of iron and the total mass of oxygen in the mixture:
With the mass of each element we can calculate the number of moles of each atom, using the atomic weight:
0.360 g Fe * 1 mol / 55.845 g = 0.00645 moles of Fe
0.140 g O * 1 mol / 16 g = 0.00875 moles of O
The number of moles of Fe in the mixture is equal to the number of moles of FeO plus two times the number of moles of Fe₂O₃:
0.00645 = [tex]2*n_{Fe2O3} +n_{FeO}[/tex] eq A
The number of moles of O in the mixture is equal to the number of moles of FeO plus three times the number of moles of Fe₂O₃:
0.00875 = [tex]3*n_{Fe2O3} +n_{FeO}[/tex] eq B
So now we have a system of two equations and two unknowns, we solve for [tex]n_{Fe2O3}[/tex]:
From eq A:
[tex]n_{FeO3}=0.00645-2*n_{Fe2O3}[/tex]
Replacing in eq B:
[tex]0.00875=3*n_{Fe2O3} + (0.00645-2*n_{Fe2O3})\\0.00230=n_{Fe2O3}[/tex]
Now we just need to convert moles of Fe₂O₃ into grams, using the molecular weight:
0.00230 moles * 159.66 g/mol = 0.367 g Fe₂O₃
