Sulfite ion, SO3^(2-), can be oxidized to SO4^(2-) using hydrogen peroxide. A 0.01610 g sample, which was 32.41 % sulfite, was treated with hydrogen peroxide and a solution of BaCl2. How many grams of BaSO4 should be produced?
2007-01-25 02:49:52 · 2 answers · asked by Alan l 1 in Science & Mathematics ➔ Chemistry
.01610g * .3241 * mole SO3-2/80g = .0006523 moles
0006523 mole * (233.43 g BaSO4/mole) = .1522 g BaSO4
2007-01-25 03:04:11 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
Isn't stoichiometry fun?
OK...First, write out an equation for the reactions:
SO32- --> SO42- + Ba2+(aq) --> BaSO4(s)
You can see that 1 mol of sulfite ion will lead to 1 mole of BaSO4 being formed (This obviously isn't a complete equation, but it shows everything you need).
Now, you start with the mass of the sample, and recognize that only 32.41% of it is sulfite ions. So, you actually began with 0.005218 grams of sulfite ions. Convert that to moles by dividing by the molar mass of an SO32- ion, and you have the moles of BaSO4(s) that you'll form. Convert that back to grams using the molar mass of BaSO4, and you're done.
2007-01-25 11:02:41 · answer #2 · answered by hcbiochem 7 · 0⤊ 0⤋