solitanze - 4-2-2016 at 16:33
A magnesite ore contains 80 wt% MgCO3, 15 wt% SiO2 and 5 wt% H2O.
This ore is fed to a treatment unit where some of the silica and all of the water is removed as a waste stream. The purified ore now contains 98 wt%
MgCO3, 2 wt% SiO2 . This mixture is sent to a furnace where it is calcined.
The reaction is:
MgCO3 —> MgO + CO2
Some of the MgO formed reacts with SiO2 in the side reaction
MgO + SiO2 —> MgSiO3
and no SiO2 remains after calcination.
The mixture then passes to a reactor where it reacts with chlorine (Cl2) and cokes (C):
MgO + C + Cl2 —> MgCl2 + CO
MgSiO3 doesn’t react. Cl2 is supplied 50% in excess of the stoichiometric amount (the amount needed to convert all the MgO). C is supplied in 100%
excess.
Calculate
The amount of purified ore and the molar composition
The amount of waste produced (SiO2 + H2O) + composition
The amount of magnesite ore to produce 1000 kg of MgCl2.
I really hope this is clear because English is only a second language for me.
Anyway, I figured calculating the amount of required MgCl2 is the best way to solve this problem, and then proceed from there.
1000 kg MgCl2 —> 10 501 mol MgCl2 —> 10 501 mol Cl2 & C required
Cl2 and C are in excess: 15 751 mol Cl2 + 21 002 mol C required
This was the easy part.
Obviously 10 501 mol MgCl2 requires 10 501 mol MgO. But this is where I can’t proceed. The feed to reactor also contains MgSiO3 (from MgO + SiO2
—> MgSiO3)
so more MgO is needed.
Using an overall balance for the process didn’t solve much (because the amount of CO2 produced in the furnace is unknown).
Any help at all would be greatly appreciated.
Thanks in advance.
Magpie - 4-2-2016 at 16:48
I don't see how the first 2 questions can be solved as stated in that order as no basis weight is given.
If the statement of the 3rd question, ie, 1000 kg of MgCl2, is used as a basis then all 3 questions can be solved.
blogfast25 - 4-2-2016 at 18:57
Ooopsie. I read Mn, not Mg.
[Edited on 5-2-2016 by blogfast25]
gsd - 4-2-2016 at 19:41
Basis: 1000 kg of Ore
80 % MgCO3 = 800 kg of MgCO3
15 % SiO2 = 150 kg of SiO2
5 % H2O = 50 kg of H2O
Now after waving some magic wand we get "purified" ore which contains 98 % MgCO3 and 2 % SiO2.
Since all of original MgCO3 remains intact, we have the purified ore which now weighs 800/0.98 = 816.33 kg and the new ore composition is:
MgCO3 = 800 kg
SiO2 = 16.33 kg
This purified ore is now calcined:
1)MgCO3 -------------> MgO + CO2
84.3 -------------> 40.3 + 44
2)MgO + SiO2 -------------> MgSiO3
40.3 + 60 -------------> 100.3
Now 800 kg of MgCO3 will give (40.3 X 800)/84.3 = 382.44 kg of MgO
The 417.56 kg of CO2 produced will escape.
The 16.33 kg of SiO2 of purified ore will react as per reaction 2 taking away with it (16.33 X 40.3)/60 = 10.96 kg of MgO
to produce (16.33 X 100.3) / 60 = 27.3 kg of MgSiO3
The new composition of this treated ore now is:
MgO : 382.44 - 10.96 = 371.48 kg
MgSiO3: = 27.3 kg
This Ore is now treated with Carbon and Cl2 to produce MgCl2:
3)MgO + C + Cl2 ----------> MgCl2 + CO
40.3 + 12 + 70.9 -----------> 95.2 + 28
So 371.48 kg of MgO will need (371.48 X 12)/ 40.3 = 110.6 kg of Carbon
But carbon used is 100 % excess, so actual C used is 221.2 kg and C remaining in the mixture is 110.6 kg
Similarly 371.48 kg of MgO will need (371.48 X 70.9)/ 40.3 = 653.55 kg of Cl2. But Cl2 used is 50 % excess, so actual Cl2 used is 1.5 X 653.55 =
980.32 kg out of which 326.77 will escape along with the CO formed.
MgCl2 produced will be: (371.48 X 95.2)/ 40.3 = 877.54 kg
With this calculations as your base you can now get you answers.
gsd
PS: I don't know where and in which year you are studying your Chem Engg. But the book "Chemical Process Principles vol 1" by Hougen & Watson will
be extremely useful to you.
gsd
[Edited on 5-2-2016 by gsd]
gsd - 4-2-2016 at 21:40
Here is a link for Hougen & Watson - Chemical Process Principles
Vol 1: Material and Energy Balances
Vol 2: Thermodynamics
Vol 3: Kinetics and Catalysis
All three volumes combined - http://krishikosh.egranth.ac.in/bitstream/1/2023593/1/BPT108...
gsd
solitanze - 23-2-2016 at 05:29
Thank you very much for commenting. It's been a while but I was in the hospital for a while (appendicitis).
I eventually solved the problem on my own, here's what I did (could be useful/interesting for someone so I will post it anyway):
I again started from 1000 kg of magnesium chloride as a basis, because this was emphasized in the original exercise.
I already knew that 10 501 mol MgCl2 requires 10 501 mol MgO theoretically, or 10 501 mol MgCO3 in the concentrated mixture. But a little more is
needed because some of it reacts with the silica in the concentrate.
Suppose x is the amount of MgCO3 (in grams) that gets converted to MgSiO3.
Or in moles: x / 84.33 (84.33 = molar mass M, in g/mol)
This amount reacts with the SiO2. So there must also be present ( x / 84.33 ) mol SiO2 in the mixture.
Or in grams: ( x / 84.33 )* 60.06 g SiO2 (60.06 g/mol = molar mass of SiO2)
10 501 mol MgCO3 --> 885 540,27 g MgCO3
The total mass of concentrated mixture is now: (x / 84.33*60.06 + x + 885 540.27) g
This mixture contains 2W% of SiO2:
( ( x / 84.33 )* 60.06) / ( x / 84.33*60.06 + x + 885 540.27 ) = 0.02
x = 26 123,75 g MgCO3
It follows that the concentrated mixture contains 911 664,02 g of MgCO3 and 18 605,39 g of SiO2.
From here we can easily calculate all else that is needed. The amount of magnesite ore required is 1 140 kg.
However, gsd's method is easier. Why didn't I think of that? 
I plugged all the different values in an Excel file so I could easily work with different amounts and it all checks out. All in all very
interesting...
Gsd, I'm in my second year of chemical engineering at the university of Antwerp.
The pdf you provided looks indeed very useful, I will definitely check it out.
Have a nice day.