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chornedsnorkack
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Quote: Originally posted by quantumcorespacealchemyst  | No, the first one, its 2H2SO4 but there is only 1HSO4- and H2O. leaving
4 hydrogens on the left, 3 on the right
that equation is from Wikipedia and needs correction. I don't know the correct mechanism.
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In this case, you would also have 2 sulphurs on the left, 1 on the right.
The equation would balance if number 2 applied to entire MnO3+HSO4-, not just MnO3+. Could the equation be presented in a way that is ambiguous or
incorrect?
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quantumcorespacealchemyst
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i guess it may be Mn2O7 + H2SO4 ---> H+, MnO4-, MnO3+, HSO4-
also, while i am unsure now, it seems the mixture lightened a tiny amount. i am unsure what it is. as far as i know its is from an interaction of
Potassium permanganate KMnO4,
Terbium sulfate Tb2(SO4)3,
sulphuric acid H2SO4,
and Hydrochloric acid HCl.
there was a green precipitate (with some brownish (MnO2?) it seemed) that was insoluble most of the time until addition of HCl. i read HCl acts on
MnO2 to Make the chloride (III) but it evolves Cl2, which wasn't observed. also though, the contact of MnCl3 with Cl- makes different negative ions of
Manganese chlorides, although this would need the Cl2 made in the first place. the flask was in snow and it is possible, i think that the cl2 was
dissolved as it was made back to HCl and HOCl, itself possibly decayed to HCl, so this seems possible. also this put the green stuff, the Mn2O7, in
solution.
also the amount of HCl added did not seem sufficient to act on the Mn2O7/HMnO4.
[Edited on 29-1-2015 by quantumcorespacealchemyst]
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learner1112
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mnso4 industrial route
mso4 industrial route
i am trying to figure out an economical industrial route of processing mnso4 from natural ore. essentially mno2. the first process is to burn sulphur
and inject the so2 formed into the crushed ore and water solution. the slurry thus obtained is kept at ph4-4.5 by adding lime. after the reaction the
slurry is filtered and the liquid which is pink in colour is obtained. now the problem is how to seperate the mnso4 present in the solution. can
anyone guide me, as when i try to heat the solution its ph goes down and it becomes acidic with evolution of so2 gas and probably the mnso4 is
converted baqck to mno2
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blogfast25
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It would help if you quit with the 'txtinglish' and started using proper chemical symbols.
But you seem to have it all wrapped up: SO2 indeed reduces a slurry of MnO2 to MnSO4. Filter off insoluble/residue and you have a solution of crude
MnSO4. This solution will only be faintly pink if it is very concentrated.
Your pH goes down mainly because you're removing excess SO2. It will not revert back to MnO2 upon concentrating.
Evaporate to dryness to obtain MnSO4.H2O, which is faintly pink. But your product thus obtained is likely to be contaminated with iron though and thus
darker in colour.
Keeping the pH at 4/4.5 during the SO2 addition step could help keeping most of the iron as insoluble Fe(OH)3 and reduce the level of iron
contamination in the final product.
[Edited on 9-2-2015 by blogfast25]
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quantumcorespacealchemyst
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MnO2 + H2SO4 ---> MnSO4 + H2O + O2
I read that MnO2 + H2SO4 ---> MnSO4 + H2O + 1/2O2
Is this correct? If so, why not pulverize the ore, and react the MnO2 from Mn+4 to Mn+2 with sulfuric acid turning O-2 to 1/2O2, filter and evaporate?
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quantumcorespacealchemyst
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MnO2 + H2SO4 ---> MnSO4 + H2O + O2
If there is SO2 it may be S+4 turning to S+6, giving 2e- to Mn+4 (MnO2), making Mn+2 (MnO) with the O-2 attaching to (now) SO2+2--->SO3. Ideally,
SO3 + H2O---> H2SO4, which reacts with MnO forming MnSO4 and H2O.
[Edited on 9-2-2015 by quantumcorespacealchemyst]
With the lime, I am unsure what it is exactly, if it is CaO, Ca(OH)2, carbonates or with different beginning parts, Al, Mg, or other stuff.
If they switch anion/cations and are all soluble, I don't know which ones are present in salt crystals and or if they double salt. I recently asked a
teacher I had learned from in school a similar question about the nature of precipitating soluble ion solutions and was told that the "activity
coefficient" of the ions are needed and some complex mathematics, the "specific ion theory" being what is needed, it seems (http://en.wikipedia.org/wiki/Specific_ion_interaction_theory).
in this case if using Calcium salts, it seems that the low solubility of Calcium hydroxide and especially Calcium carbonate can be used to separate
concentrated solutions of MnSO4. I don't know how pure it can be made by that technique.
I hope some people can clarify, verify all this. I haven't done this yet. it seems interesting.
I thought the lime was unnecessary, and then I read what you wrote about SO2 being driven out of solution on heating and now I am puzzled too. Does
this happen when you don't add lime? I don't know how soluble SO2 is in water, and if the SO2 is from boiling it out of solution, then I don't know
why the pH lowers. It seems odd if it is reversal of Mn+2 as if the 2e- (Mn+2 -> Mn+4 +2e-) comes from the Mn+2, where does it go? if the H2SO4
somehow turns to H2SO3 (S+6 -> S+4), turning to SO2 and H2O with heating, the pH lowering seems odd. I don't know, I hope someone here does.
[Edited on 9-2-2015 by quantumcorespacealchemyst]
[Edited on 9-2-2015 by quantumcorespacealchemyst]
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blogfast25
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quantumcorespacealchemyst:
It's hard to respond to your garbled junk because although it contains truth it also contains stuff you clearly haven't digested very well, a constant
in your posts. I suggest very strongly to start your career as a chemical experimenter by cutting your teeth on simple and satisfying experiments,
rather than constantly over-reaching and ending up very confused.
I believe the lime (which would be Ca(OH)2 in this case) is there mainly to control pH. Mn bearing ores often contain iron which you want to keep out
of your end-product.
Fe(OH)3 is very, very insoluble, requiring quite low pH to enter solution. Keeping pH at 4 to 4.5 during leaching with the SO2 would keep the iron
insoluble, or at least that's the idea.
In contrast, using actual H2SO4 would definitely dissolve all or part of the iron.
[Edited on 9-2-2015 by blogfast25]
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learner1112
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exactly the lime is used to controll the ph. and so2 is used instead of h2so4 because that way it is cheaper.
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quantumcorespacealchemyst
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why does it give off SO2 on heating with pH lowering?
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blogfast25
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During the reactive leaching, SO2 is absorbed by the lime in the slurry to form calcium sulphite (CaSO<sub>3</sub> and further down the road calcium bisulphite (Ca(HSO<sub>3</sub><sub>2</sub>. This combination forms the (primitive) buffer system (pH controller). Once pH is below 5 (or so) the reduction of the MnO2 starts and
MnSO4 starts forming and solubilising.
When the reactive leaching step has dissolved all (or most anyway) of the manganese the slurry will still reek of excess SO2. After filtering, this
excess sulphite will partly oxidise (with air) to sulphate, part of it will simply escape. Nothing mysterious here...
Come to think of it, it's probably recommended to blow air through the slurry prior to filtering, to flush out the excess SO2 and/or oxidise it to
sulphate.
[Edited on 10-2-2015 by blogfast25]
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quantumcorespacealchemyst
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Surprise~
I checked it yesterday, it turns out 14 days exactly since the red/brown picture.
[Edited on 10-2-2015 by quantumcorespacealchemyst]
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blogfast25
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It would help of course if we knew what the picture was and what exactly you did to get that solution.
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quantumcorespacealchemyst
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1st page
H2SO4, Terbium, KMnO4,
then HCl(impurity?it was cloudy)
the pictures of the flask show the green/yellow insolubles.
HCl put it all in solution. didn't see bubbles.
darkyellow/lightbrown color, darkened alot overnight, and after.
picture of that on bottom of 1st page.
14 days, checked it and took a picture^^^ very light/clear
the HCl was made by gassing dilute H2O2 with HCl from a reaction. it sat a long time and possibly absorbed acetyl chloride and possibly some
1,4-dibromoacetophenone.
it was about a quarter it's original volume (the H2O2) when removed from the reaction vessel about a month later. it was cloudy colorless and
decomposed baking soda.
[Edited on 10-2-2015 by quantumcorespacealchemyst]
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chornedsnorkack
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| Quote: Originally posted by quantumcorespacealchemyst | I read that MnO2 + H2SO4 ---> MnSO4 + H2O + 1/2O2
Is this correct? If so, why not pulverize the ore, and react the MnO2 from Mn+4 to Mn+2 with sulfuric acid turning O-2 to 1/2O2, filter and evaporate?
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1) Oxidizing molecular oxygen is a difficult process, unless the oxidant is extremely strong, like ferrates. MnO2 reaction with H2SO4 would be
sluggish
2) SO2 is a cheap substance. Why spend money to oxidize it to pure H2SO4 and then discard the oxygen again, if you can just use SO2 directly to reduce
MnO2 much faster?
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quantumcorespacealchemyst
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yea
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blogfast25
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| Quote: Originally posted by chornedsnorkack |
1) Oxidizing molecular oxygen is a difficult process, unless the oxidant is extremely strong, like ferrates. MnO2 reaction with H2SO4 would be
sluggish
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Oxidising molecular oxygen (O<sub>2</sub> cannot be oxidised. The only
real oxidation states for oxygen are -2 and -1. There are NO positive oxidation states for oxygen.
Going from molecular oxygen (oxidation state 0) to -1 or -2 is a reduction, not an oxidation. Throwing ferrate at it doesn't change that.
[Edited on 22-2-2015 by blogfast25]
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j_sum1
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| Quote: Originally posted by blogfast25 | | Oxidising molecular oxygen (O<sub>2</sub> cannot be oxidised. The only
real oxidation states for oxygen are -2 and -1. There are NO positive oxidation states for oxygen. |
FOOF???
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blogfast25
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C'mon j_sum1. You're right of course but PLEASE look at context. We can also 'oxidise' O to +3 by stripping it of 3 electrons in a machine but on
meeting any other atom or ion that state would immediately be reduced. And that's not what he was talking about.
Even in O2F2, which is highly unstable, the bonds are more covalent and clearly unstable.
I'm guessing he ['chorned'] simply got a bit confused between oxidation and reduction.
[Edited on 23-2-2015 by blogfast25]
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chornedsnorkack
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| Quote: Originally posted by blogfast25 |
I'm guessing he ['chorned'] simply got a bit confused between oxidation and reduction.
[Edited on 23-2-2015 by blogfast25] |
No - forgot to spell out clearly oxidation TO molecular oxygen.
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j_sum1
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| Quote: Originally posted by blogfast25 |
C'mon j_sum1. You're right of course but PLEASE look at context. We can also 'oxidise' O to +3 by stripping it of 3 electrons in a machine but on
meeting any other atom or ion that state would immediately be reduced. And that's not what he was talking about.
Even in O2F2, which is highly unstable, the bonds are more covalent and clearly unstable.
I'm guessing he ['chorned'] simply got a bit confused between oxidation and reduction.
[Edited on 23-2-2015 by blogfast25] |
I find the mere concept of FOOF hilarious -- it is just so extreme. And
that article never fails to make me laugh. Satan's kimchi indeed.
In any case your assessment of the situation stands. You specified "real" oxidation states of Oxygen. Playing with FOOF is so far out in imagination
land that I think it is unlikely to be something real chemists would ever really encounter.
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quantumcorespacealchemyst
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it tuns out, the H2SO4 i was using was 2N, nowhere near the 98% o 95% i thought i had (wrote one o those somewhere). that explains my constant
puzzlement about it's peculia properties and failed reactions.
the reaction here evaporated on standing over time leaving a dark green oil, presumably KMn2O7 which i brought outside to dunk in a coldwater
container and then added a little KOH.
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