% Composition : GChem

[IWantSmile]

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Dang~! I even have a hard time to understand this question.
Can someone help me? Please!

Question: A taconite ore consisted of 35% Fe3O4 and the balance siliceous impurities. How many tons of the ore must be processed in order to recover a ton of metallic iron if there is only 75% recovery ?


Thanks in advance 🙂

Oh! the answer is 5.27 tons.

 

[IWantSmile]

.....???

 

[HurricaneKatt]

...Tons? I have no clue...I don't remember how to do this. Figure out the ratio first maybe. Sorry.

 

[keliao]

i dont think the DAT wil lhave us calculating tons type of question!!

 

[Dmitry Malayev]

This is way too hard for the DAT. I don't even know the conversion factor for tons! 🙁

 

[supernova937]

Fe3O4 MW
Fe : 3*55.8 = 167.4
O: 4*16 = 64

Fe3O4 = 231.4 g/mol
%Fe = 167.4 / 231.4 = 72.34

Since the ore contains 35% of Fe3O4 and the recovery is 75%, for every ton of ore you get (0.7234)(0.35)(0.75) = 0.1899 ton of metallic iron.

To get 1 ton of iron = 1 / 0.1899 = 5.265 ton of ore

 

[waystinthyme]

you won't have any questions this hard, just make sure you get down the basic calculations and you should be fine.

-waystinthyme

 

[IWantSmile]

Fe3O4 MW
Fe : 3*55.8 = 167.4
O: 4*16 = 64

Fe3O4 = 231.4 g/mol
%Fe = 167.4 / 231.4 = 72.34

Since the ore contains 35% of Fe3O4 and the recovery is 75%, for every ton of ore you get (0.7234)(0.35)(0.75) = 0.1899 ton of metallic iron.

To get 1 ton of iron = 1 / 0.1899 = 5.265 ton of ore

THANK YOU for your great help, "supernova".

 

[IWantSmile]

...Tons? I have no clue...I don't remember how to do this. Figure out the ratio first maybe. Sorry.

Thanks,