Question

A mixture of Fe2O3 and FeO was found to contain 72.00% Fe

bymass. What is the mass of Fe2O3 in 0.500 g of this mixture...how
doi work this..i dont even know what to look for as the given
asneeded?

Answer 1

The mass of Fe₂O₃ in 0.500 g of mixture is 0.367 g.

Step-by-step 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:

• 0.500 g * 0.72 = 0.360 g of Fe
• 0.500 g * 0.28 = 0.140 g of O

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 =
`2*n_(Fe2O3) +n_(FeO)` 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 =
`3*n_(Fe2O3) +n_(FeO)` eq B

So now we have a system of two equations and two unknowns, we solve for
`n_(Fe2O3)`:

From eq A:

`n_(FeO3)=0.00645-2*n_(Fe2O3)`

Replacing in eq B:

`0.00875=3*n_(Fe2O3) + (0.00645-2*n_(Fe2O3))\\0.00230=n_(Fe2O3)`

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