Ashendale - 11-1-2006 at 12:51
Ok, I've got an exercise that I solved, but as I need to send it to way for checking, I need to make sure I solved it correctly.
The text itself is simple
"0,654 tonnes of FeS2 with 13,8% other compounds, was used to manufacture H2SO4. The amount of H2SO4 manufactured was 46% 960 dm3 (1,357 g/cm3). Find
the yield of sulfuric acid (In %)
The problem itself is, I get a really low yield (6,5%) and I'm not sure if it's right. Here's my solution:
4FeS2 + 11O2 -> 2Fe2O3 + 8SO2
2SO2 + O2 -> 2SO3
SO3 + H2O -> H2SO4
m(FeS2+o.compound) = 654kg * 0,138 = ~90,3kg
m(FeS2) = 563,7kg = ~564kg
n(FeS2) = 564kg / 120g/mol = ~4700mol
Theoretical amount produced will follow:
n(SO2) = n(FeS2)*2
n(SO2) = 9400mol
n(SO3) = n(SO2)
n(SO3) = 9400mol
n(SO3) = n(H2SO4)
n(H2SO4) = 9400mol
m(H2SO4 ) = 9400mol*98,09g/mol = ~920kg
Theoretical part ends
m(solution) = 960dm3 * 1357g/cm3 = ~130kg
m(H2SO4) = 130kg * 46,0% = ~59,8kg
%(yield) = 59,8kg / 920kg * 100% = 6,5%
Did I solve it right? :S
[Edited on 11-1-2006 by Ashendale]
[Edited on 11-1-2006 by Ashendale]
Magpie - 11-1-2006 at 14:21
I think you slipped a decimal point. Should be 1300 kg vs 130 kg.
Shortcuts: Work in kg-moles. There is nothing sacred about g-moles. Go directly to kg-moles H2SO4 = kg-moles FeS2 due to stoichiometry. 
neutrino - 11-1-2006 at 14:36
I did this problem a slightly different way.
0.654 tons (metric, I assume) = 654000 grams ore
654000 grams ore = 560300 grams FeS<sub>2</sub>
560300 grams FeS<sub>2</sub> comes to 9340 moles of sulfur
960L = 960000mL
960000mL = 1302720 grams of acid solution = 600000 grams of H<sub>2</sub>SO<sub>4</sub>
600000 grams of H<sub>2</sub>SO<sub>4</sub> = 5920 moles of sulfur.
In your reaction, every mole of sulfur produces a mole of acid. Thus, we can divide moles of sulfur in the product by moles of sulfur in the reactant
and we get a yield of 63.4%.
I hope I didn't make a silly error anywhere in there.
Ashendale - 12-1-2006 at 05:30
Thanks people 
Solved it and sent it, i'll get answers in a week so I'll tell you if they were correct