I have a question that I do not how to solve it even with the answer from the solution manual. Can someone please explain to me on how to solve it. Thank You. (Happy New Year!)
Q: A 2.00 g mixture of Fe2O3 and impurities having no iron, is analyed as containing 1.25 g of Fe. What is the percent of Fe2O3 in the mixture?
A: 89.3%
2006-12-31 06:37:50 · 1 answers · asked by AUNIDLY 1 in Science & Mathematics ➔ Chemistry
Atomic weights: Fe= 55.8 O = 16 Fe2O3 = 160 (to three significant figures)
1.25gFe x 1gatomFe/55.8gFe x 1molFe2O3/2gatomFe x 160gFe2O3/1molFe2O3 = (1.25)(160)/(55.8)(2) = 1.79gFe2O3
1.79g/2.00 x 100% = 89.5%
To find the amount of Fe2O3 corresponding to 1.25gFe: Multiply by the first factor, which comes from the atomic weight. g Fe cancel, leaving gram-atoms Fe. The second factor comes from the formulas of Fe2O3 and Fe. The gram-atoms Fe cancel, leaving moles Fe2O3. The third factor comes from the formula weight of Fe2O3. The molFe2O3 cancel, leaving g Fe2O3.
2006-12-31 06:48:50 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋