Amixture of fe2o3 and feo was found to contain 72.00% fe
bymass. what is the mass of fe2o3 in 0.500 g of this
doi work this..i dont even know what to look for as the given
asneeded?

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 =             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 =               eq B

So now we have a system of two equations and two unknowns, we solve for :

From eq A:

Replacing in  eq B:

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

i believe it is c.) 1575 j.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%