The molar mass of Fe2(SO4)3 is 399.91 g/mol
2006-12-22 01:44:06 · 2 answers · asked by sydney 2 in Science & Mathematics ➔ Chemistry
399.91 g Fe2(SO4)3 contains 288 g SO42-
288 g SO42- is present in ------399. 1 g Fe2(SO4)3
hence 3.471024g SO42- is present in ---
(399.1*3.471024)/288 g Fe2(SO4)3
i.e. 4.8 g Fe2(SO4)3
2006-12-22 02:13:56 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The mass of a sulfate ion is (32.07 + 4x16.00) is 96.07g. There are 3 sulfates in each molecule so their mass would be c x 96 = 288.210g.
So, your conversion factor is 299.21g sulfate for every 399.91 g of Fe2(SO4)3.
3.471024 g sulfate x 399.91g ferric sulfate/299.21 g sulfate =
4.6392 g
2006-12-22 18:09:45 · answer #2 · answered by The Old Professor 5 · 0⤊ 0⤋