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Ashendale
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Mood: Hydrated posted on 11-1-2006 at 12:51
Exercise (solved but need to verify it's correct)

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
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Mood: Chemistry: the subtle science. posted on 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. The single most important condition for a successful synthesis is good mixing - Nicodem
neutrino
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Mood: oscillating posted on 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
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Mood: Hydrated posted on 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 Sciencemadness Discussion Board » Fundamentals » Beginnings » Exercise (solved but need to verify it's correct) Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues 