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Mabus
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Quote: Originally posted by S.C. Wack  | | The urea method does not use hydroxides esp. KOH w/10% water. It requires the most anhydrous carbonate you got. Same with the ferrocyanide.
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True, both work and both are very OTC. I too would definitely want some ferro or ferricyanide, since there are fewer side products. But alas, gotta do
what I can with what I got.
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AJKOER
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In my opinion, an interesting path to CN- from OCN- based on some radical chemistry, which I argue is the basis for a famous patent, see Clancy Patent
discussion at http://www.sciencemadness.org/talk/viewthread.php?tid=87030#... .
Or, disregard my theoretical discussion and lookup details on execution of the patent, which employs an electrolysis approach.
[Edited on 21-1-2019 by AJKOER]
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S.C. Wack
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Inorg Syn says if done right the potassium cyanate is nearly pure in high yield and the Na salt is easily purified...btw with carbonate and
ferrocyanide the extra mol of cyanide product comes with one of cyanate. It might be wiser to add nothing at all to the ferrocyanide and lose N2 than
to heat it with NaOH.
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PirateDocBrown
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Fe7(CN)18 + H3PO4 = FePO4 + HCN
That's laundry bluing, you can get it at Wal-Mart. Dry it, first.
Condense the HCN with a very cold condenser fluid, or you'll lose a lot to evap. Catch the condensate in chilled EtOH, the solution is much easier to
handle, Pure prussic acid boils around RT.
Make damn sure you have good ventilation, duh.
You want Na or K CN, dissolve the base in the EtOH first. Even easier.
[Edited on 2/11/19 by PirateDocBrown]
Phlogiston manufacturer/supplier.
For all your phlogiston needs.
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Fantasma4500
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2. The process of Claim l wherein a) step for mixing of raw materials is characterized in that 2.5~2.8:i mol ratio of urea and sodium carbonate are
charged the blender, and 200 mesh-sized iron powder and 200 mesh-sized anthracite coal powder mixed by 1:1 weight ratio are added by the quantity
corresponding to 0.2-0.4 % of raw material, and then raw materials and additives are uniformly mixed.
3. The process of Claim 1 wherein b) step for first-order reaction is characterized in that the mixture of claim 2 is charged a reactor with and then
the temperature of the reactor inside is hold at 300-350 °C for 2 hours to make the reactants of the mixture react.
4. The process of Claim 1 wherein c) step for second-order reaction is characterized in that the temperature of the reactor inside is hold at 650-750
°C for 1 hour to make the first-order reaction product deoxidized.
https://patents.google.com/patent/WO2016199944A1/en
i suppose butane could be pumped into reaction vessel to keep oxygen out, its very heavy.
also RIP anders
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macckone
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So that patent appears to be making sodium cyanurate then heating with iron and carbon to catalyze to sodium cyanide.
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walruslover69
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I have carried out a similar procedure by nurdrage with success a long time ago. Carbon does the reducing. what is the role of iron? is it a catalyst?
what would that mechanism be?
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Fantasma4500
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i just realize, as with NaNO3 to NaNO2, iron oxalate can be used as it thermally decomposes at relatively low temperature, forming superfine iron
powder
this would keep oxygen out and at the same time reduce
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macckone
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walruslover69,
the carbon and iron appear to be strictly catalytic as they are only a few percent of the other reactants.
I believe they are using the urea itself as a reductant. It is added in a good molar excess.
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walruslover69
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I didn't see that It was only a catalytic amount of iron and carbon. Makes sense that the iron would be an oxygen scavenger. I don't think urea is the
reductant though, Is it truly just a thermal decomposition then?
The first heating step with urea and sodium carbonate at 350 forms sodium cyanate. The urea decomposes, and it's quite clear when the reaction is over
and stops giving off large amounts of ammonia.
The second step is the conversion of cyanate to cyanide. I've read through several papers and threads and I'm still confused about the actual process
going on. I always assumed it was a reduction of NaCNO + C ----> NaCN + CO or 2NaCNO + C ----> 2NaCN CO2. But if the reaction takes place
without a direct reduction that would mean that NaCN is the more thermodynamically stable compound and only a catalyst is needed.
doing first order calculations from wiki's thermo data
NaCNO enthalpy formation = -400kj/mol
entropy = 119.2 j/mol
NaCN enthalpy formation = -91kj/mol
entropy= 115.7 j/mol
O2 entropy = 205,14 j/mol
For the reaction 2NaCNO --> 2NaCN + O2
Gibbs free energy = 560 kj/mol at 25C and 425 kj/mol at 700C.
these values are for room temp, so one problem might be that the thermo data is significantly different in the molten phase, but I couldn't find data
for that. There is the kinetic factor of O2 constantly evolving into open air that would shift the equilibrium to cyanide, but I would be skeptical
that it would be enough to could overcome the 425 kj/mol at any significant rate.
Am I missing something ?
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macckone
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walruslover69,
There is excess urea, it will form biuret and cyanuric acid giving off ammonia. But there is not enough sodium carbonate for it to react with so it
is going to act as a reducing agent at the higher temperature. The ratio is 2.8 moles of urea to 1 mole of sodium carbonate. That means there is 0.8
moles excess to act as a reducing agent. With the iron and carbon catalyst, that may be sufficient.
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walruslover69
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The urea will just thermally decompose. I believe it starts around 100C There is no way it would be stable at 350C for 2 hours.
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