A 0.0530 g sample of iron sulfate was dissolved in dilute HCl and treated 10% BaCl2 solution. The BaSO4 precipitate weighed 0.0928 g. Determine the formula of this sulfate of iron.
2007-02-06 18:56:36 · 2 answers · asked by Alan l 1 in Science & Mathematics ➔ Chemistry
The answer is Fe2(SO4)3.
When sulfate forms a compound with iron, there can be two possibilities. It could either from FeSO4 or Fe2(SO4)3.
To determine which iron sulfate formed in this case you must first determine the amount of sulfate in 0.0928g of BaSO4.
0.0530g BaSO4 (1mol BaSO4 / 233.43g BaSO4) (1mol SO4--/1mol BaSO4) = 3.975 x 10^-4 mol SO4--.
This calculation shows that in the 0.0928 g of Barium Sulfate Precipitate, 3.975 x 10^-4 mol of sulfate ion is included.
This amount of sulfate can only come from 0.0530g Fe2(SO4)3.
0.0530g Fe2(SO4)3 (1mol Fe2(SO4)3 / 399.87g Fe2(SO4)3) (3mol SO4--/1mol Fe2(SO4)3) = 3.975 x 10^-4 mol SO4--.
If you do the calculation for FeSO4, you'll see that it 0.0530g of it can only have 3.488x10^-4 mol SO4--, thus making it Impossible to form 0.0928g of BaSO4 ppt.
Email me if you want more explanation. Good luck in your Chem.
2007-02-06 21:59:24 · answer #1 · answered by †ђ!ηK †αηK² 6 · 0⤊ 0⤋
formula of sulphate is always S04 dear
the question must be something else
2007-02-07 03:17:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋