Actually, the mass of Fe in the residue is 2.407-Al2O3. The Fe2O3 has all lost its oxygen, leaving only Fe.
So, if you started out with x moles of Fe2O3 and y grams of Al2O3, then converting the Fe2O3 to grams,
159.69x + y = 3.064
55.85(2x) + y = 2.407
The equations may look a bit funny, but we really don't have to convert the moles of Al2O3 to grams, since the mass doesn't change. And, of course the (2x) appears because each mole of Fe2O3 has 2 Fe atoms in it.
Solving the equations yields
x = 0.01369 moles Fe2O3
y = 0.8778 grams of Al2O3
Now we need to convert the moles back to grams: .01369 moles = 2.1862 grams Fe2O3
2.1862/3.064 = 71.35%
A sample weighing 3.064 g is a mixture of Fe2O3 (molar mass = 159.69) and Al2O3 (molar mass = 101.96).
When heat and a stream of H2 gas is applied to the sample, the Fe2O3 reacts to form metallic Fe and H2O(g). The Al2O3 does not react.
If the sample residue weighs 2.407 g, what is the mass fraction of Fe2O3 in the original sample?
I am not exactly sure where to start with this question. Would this be correct?
Mass of Fe2O3 = [(2.407 g - mass of Al2O3) / 55.85 g/mol] x (1 mol Fe2O3 / 2 mol Fe) x 159.69 g/mol
2 answers
Actually, we really didn't have to convert the moles back to grams, since we had the mass of Al2O3 already. Just subtract its fraction from 100%.
2 answers