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AJKOER
Radically Dubious
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The oxalate approach apparently works. However, here are my recommendations:
> If exploring the chemistry of Oxalic acid, the dihydrate acid is possibly safer to work with as Oxalic acid can produce a vigorous decomposition
reaction with strong oxidizers.
> If trying to make concentrated acid (H2SO4, H3PO4,...) avoid all hydrates whether in the form of the Oxalic acid or the hydrated acid salts. This
is because, unlike Cl2/SO2/H2O reaction, water is not consumed.
> Do not discard the precipitated Oxalate salts. For example, Iron Oxalate can be used to form a very fine form of Iron powder. Nano like crystals
are obtained with the decomposition of other Oxalates as well.
> Shop around for the best deal on H2C2O4.
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Lithium
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if the oxidant used for oxidising H2SO3 to H2SO4 is HOCl, then would TCCA, trichloroisocyanuric acid, be used?
(CNOCl)3 + 3H2O ---> (CHNO)3 + 3HOCl
HOCl + H2SO3 ---> H2SO4 + HCl
would this reaction work?
NH2Cl + H2SO3 + H2O ---> NH4Cl + H2SO4
NH2Cl can be produced via:
(NH4)2SO4 + Ca(OCl)2 ---> 2NH2Cl + 2H2O + CaSO4
or electrolytically:
NaCl + H2O + 2e- ---> NaOCl + H2
SO2 + H2O ---> H2SO3
NaOCl + H2SO3 ---> H2SO4 + NaCl*
* NaCl is recycled back to first step
SO2 would be continually bubbled into solution.
Li
[Edited on 18-4-2012 by Lithium]
[Edited on 18-4-2012 by Lithium]
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weiming1998
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Quote: Originally posted by Lithium  | if the oxidant used for oxidising H2SO3 to H2SO4 is HOCl, then would TCCA, trichloroisocyanuric acid, be used?
(CNOCl)3 + 3H2O ---> (CHNO)3 + 3HOCl
HOCl + H2SO3 ---> H2SO4 + HCl
would this reaction work?
NH2Cl + H2SO3 + H2O ---> NH4Cl + H2SO4
NH2Cl can be produced via:
(NH4)2SO4 + Ca(OCl)2 ---> 2NH2Cl + 2H2O + CaSO4
or electrolytically:
NaCl + H2O + 2e- ---> NaOCl + H2
SO2 + H2O ---> H2SO3
NaOCl + H2SO3 ---> H2SO4 + NaCl*
* NaCl is recycled back to first step
SO2 would be continually bubbled into solution.
Li
[Edited on 18-4-2012 by Lithium]
[Edited on 18-4-2012 by Lithium] |
Here is what I think:
The trichloroisocyanuric acid idea is not very likely to work.
At first, your proposed reaction would happen, but then the formed H2SO4 and HCl reacts further with the TCCA, first forming HOCl, then Cl2,
destroying the H2SO4 formed and liberating large amounts of Cl2, which some of it reacts with more SO2, and some escapes the solution. At the end, you
would get only a small amount of H2SO4.
I don't know about the NH2Cl idea, but since Wikipedia states that sodium borohydride reduces it (probably to NH4Cl) http://en.wikipedia.org/wiki/Chloramine
there is a possibility that H2SO3 will reduce it as well.
The hypochlorite idea, like the TCCA idea, probably doesn't work as well, because something like this will occur:
First NaOCl+H2SO3==>NaCl+H2SO4 ( there might be side reactions 2NaOCl+H2SO3===>Na2SO3+2HOCl, then Na2SO3+HOCl===>Na2SO4+HCl or first HOCl
reacts with SO2, then H2SO4 reacts with Na2SO3 to form Na2SO4 and more SO2)
Then, continuing down, H2SO4+2NaOCl===>2HOCl+Na2SO4, which further oxidizes the SO2 and further reacts with the NaOCl, etc. The end product would
contain hardly any H2SO4, but a ton of Na2SO4 and NaCl.
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Lithium
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okay then, was just wondering
i just worked out, for electrolysis of CuSO4 you need 160g of it to make 50ml of pure H2SO4.
1kg CuSO4 = $30 (where i live)
3.2kg = 1L H2SO4
therefore
1L H2SO4 = $96!!!!
that's not even adding electricity bills!
so obviously CuSO4 is not viable
would an organic peroxide(AP, MEKP) oxidize the sulfurous acid to sulfuric acid?
for example:
((CH3)2C-O-O)3 + 3H2SO3 ---> 3(CH3)2C=O + H2SO4
but then, why not just use H2O2 directly
does anyone have a synth for hypochlorous acid?
Li
[Edited on 18-4-2012 by Lithium]
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weiming1998
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| Quote: Originally posted by Lithium | okay then, was just wondering
i just worked out, for electrolysis of CuSO4 you need 160g of it to make 50ml of pure H2SO4.
1kg CuSO4 = $30 (where i live)
3.2kg = 1L H2SO4
therefore
1L H2SO4 = $96!!!!
that's not even adding electricity bills!
so obviously CuSO4 is not viable
would an organic peroxide(AP, MEKP) oxidize the sulfurous acid to sulfuric acid?
for example:
((CH3)2C-O-O)3 + 3H2SO3 ---> 3(CH3)2C=O + H2SO4
but then, why not just use H2O2 directly
does anyone have a synth for hypochlorous acid?
Li
[Edited on 18-4-2012 by Lithium] |
Hypochlorous acid can be made in many different ways, but I find the most convenient way is adding Ca(ClO)2 to some vinegar. That gives about 5% HClO,
along with calcium acetate impurities. But to make sulfuric acid, you have to get rid of the calcium acetate, and the only way I can think of is
distillation, but that's a fairly dangerous job; Cl2O (from the decomposition of the HClO) can spontaneously explode on contact with organic matter!
Edit: The other ways of making HClO has salt contaminations as well, except bubbling Cl2 through water, which the Cl2 wouldn't dissolve much.
[Edited on 19-4-2012 by weiming1998]
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barley81
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Don't forget that copper sulfate is often sold as the pentahydrate, so it actually takes around 249g of it to make 98g (53mL) of pure sulfuric acid
assuming 100% yield. Hypochlorous acid is made by shaking chlorine water with mercury oxide. Read more in the SM library (this prep is from Georg
Brauer's "Handbook of Preparative Inorganic Chemistry"). Woelen suggested bubbling chlorine and sulfur dioxide into water. This will produce a mixture
of sulfuric acid and hydrochloric acid. The HCl can be boiled off and the sulfuric acid heated until it reaches a high concentration. The chlorine can
be made by adding HCl to pool chlorine (calcium hypochlorite, sodium dichloroisocyanurate, or trichloroisocyanuric acid) and the sulfur dioxide can be
made by burning sulfur (cheap), heating sulfur with sulfuric acid (oh the irony), or adding an acid to a sulfite salt. You really should try to buy
some sulfuric acid if you can, and it would be very unfortunate if you could not.
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DoctorZET
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I'm thinking about 2 methods:
first involves a tin process:
1)take some soldering alloy (Pb+Sn) and bubble the molten metals (it melts at about 170-200*C) with Cl2. Meanwhile, distill the SnCl4 gas resulted.
Pb Sn + 3Cl2 --(200*C)--> PbCl2 + SnCl4^
Now you have a quaternary stannium salt : SnCl4 (a fuming liquid at room temperature)
2)take some SnCl4 and some dehydrated CuSO4 and mix them until the mixture become almost-solid:
SnCl4 (excess) + 2CuSO4 --(time)--> Sn(SO4)2 + 2CuCl2
3)then distill the excess of SnCl4.
4)heat the remaining powder up to 200-300*C and collect the SO3 vapors:
Sn(SO4)2 --(200-300*C)--> SnO2 + 2 SO3^
(the CuCl2 impurities does no effect at all)
5)to re-use the tin in this process, heat the remaining powder (SnO2+CuCl2) with some H2 gas, in a tube, to reduce the tin(vi) oxide to tin metal and
water, copper will be also reduced to Cu and HCl :
SnO2 + 2H2 --(190-210*C)--> Sn + 2 H2O^
CuCl2 + H2 --(150-170*C)--> Cu + 2 HCl^
now you can start again to transform Sn in SnCl4 ... and so on.
Second method I asume that it consumes a lot of iron and CuSO4:
1) a boring school reaction:
Fe(powder)+CuSO4(aq) --(looong time)--> Cu(s) + FeSO4(aq)
2) Heat a bit the FeSO4:
FeSO4 --(480-500*C)--> Fe2O3 + SO3^ + SO2^
And, as you can observe, half of the CuSO4 wich was introduced in the process can not be converted into H2SO4 because of SO2 forming .
3) If somebody can tell me a easy way to reduce Fe(lll) to Fe(ll) or to Fe, this process will not consume so much iron (the iron will be recicled)
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DoctorZET
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warning: in the first process water must not be present !
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