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No, you CAN'T make sodium!
#21 20 April 2010 - 10:18 PM
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#22 1 May 2010 - 02:45 PM
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#26 12 February 2011 - 04:14 PM
TheLivingMartyr, on 12 February 2011 - 04:10 PM, said:
Yes. It is in fact the preferred method of mass scale sodium production. They use doping agents to lower the melting temperature though. However, I doubt that anyone does this at the lab scale, liquid sodium chloride is some really nasty stuff.
-Feynman Lectures on Physics II
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#27 12 February 2011 - 04:16 PM
oh wait, couldn't you reduce sodium compound with carbon monoxide?
mississippichem, on 12 February 2011 - 04:14 PM, said:
ahh yes thanks, i thought there must be some way of doing it
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#28 12 February 2011 - 04:21 PM
TheLivingMartyr, on 12 February 2011 - 04:15 PM, said:
oh wait, couldn't you reduce sodium compound with carbon monoxide?
CO isn't a strong enough reducing agent.
for the reaction
is about -2.71 V relative to the standard hydrogen electrode. One would need a really strong reducing agent as the Na metal that would form will be very reducing itself.
-Feynman Lectures on Physics II
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#30 5 October 2011 - 02:52 PM
Aluminum metal can actually reduce sodium hydroxide to sodium metal. Now I realise that many of you chemists will immediately say this is impossible, because "sodium is a more reactive element than aluminum", so here is a link showing pictures of the reaction, but with potassium being made from magnesium, instead of sodium from aluminum. http://sites.google....allic-potassium
It also works if solid NaOH is ignited with Al powder in a metal container, and a lid placed over it to prevent reaction with air. The sodium cannot be obtained in a pure state by this method, however, as it is mixed with slag. But it is still charactaristically reactive with water.
http://www.youtube.c...h?v=908rjHQ5mmc
(6)NaOH + (4)Al --> (2)Al2O3 + (6)Na + (3)H2
but note that
AlCl3 + (3)Na --> (3)NaCl + Al
It is possible to prepare sodium metal by cautiously heating sodium azide in the absence of oxygen.
(2)NaN3 --> (2)Na + (3)N2
Although lithium can burn in nitrogen, both sodium and potassium nitrides are very unstable. Sodium nitride decomposes into elemental sodium, giving off nitrogen gas, at only 87°C.
Comparing the decomposition of other metal nitrides
Similarly, the explosive decomposition of copper azide also results in the separation of the constituent elements, but this reaction happens for very different reasons.
Cu(N3)2 --> Cu + (3)N2
But the same reaction with iron (which is dangerous) will result in iron nitride.
(3)Fe(N3)2 --> Fe3N2 + (8)N2
The iron nitride can be decomposed to elemental iron and nitrogen gas above 800°C.
Fe3N2 --> (3)Fe + N2
The decomposition of calcium azide is similar to that of iron.
Ca(N3)2 decomposes above 110degC, explosively so over 140degC. The Ca3N2 that forms only decomposes at 1600degC, at which point the elemental calcium simultaneously vaporizes out with the nitrogen.
Creative Way to Make Elemental Potassium?
An idea for chemical preparation of elemental potassium, which does not require electric current. It would be impractical, but very creative. Not sure if all the reactions would work.
Ca3N2 + (6)KCl --> (3)CaCl2 + (3)K2 + N2
Distilling calcium nitride with potassium chloride in with steel-walled distillation may cause potassium to boil out. This proposed reaction would make use of Le Chatelier's principle. Although potassium boils at 759°C, it is possible that molten potassium could be produced below this temperature.
(6)CaCl2 + Ti3N4 --> (2)Ca3N2 + (3)TiCl4
The titanium nitride (m.p. 2930°C) would be crushed into a fine powder and distilled under intense heat with calcium chloride. Titanium tetrachloride (TiCl4) is a liquid which boils at only 137 °C.
(3)TiI4 + (16)NH3 --> Ti3N4 + (12)NH4I
I think titanium tetraiodide (b.p. 377 °C) could be reacted with anhydrous ammonia gas to form titanium nitride and ammonium iodide. I am not sure if the NH3 could be bubbled into molten TiI4, or if the TiI4 would need to be in the vapor phase, with the intense heat required for the reaction. The reaction would be expected to procede because TiI4 is very acidic, and because the titanium-nitrogen bonds are stronger than titanium-iodide. Wikipedia claims that TiCl4 "with ammonia, titanium nitride is formed"; this is not surprising since TiCl4 reacts with water to form titanium dioxide and hydrogen chloride.
Titanium tetraiodide melts at 150 °C. It can be prepared from easily obtainable materials:
(3) TiO2 + (4) AlI3 --> (3)TiI4 + (2)Al2O3
This post has been edited by Anders Hoveland: 5 October 2011 - 03:04 PM
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#31 22 October 2011 - 12:03 AM
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however one must realize that every thing starts in a default belief and requires a burden of proof for the default belief to be abandoned. it would not make sense for believing in positive statements' validity so the only remaining is to not believe a positive statement until
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#32 22 October 2011 - 04:45 PM
dragonstar57, on 22 October 2011 - 12:03 AM, said:
Probably only by someone able to understand what "off topic" means
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#33 22 October 2011 - 05:08 PM
John Cuthber, on 22 October 2011 - 04:45 PM, said:
This post has been edited by dragonstar57: 22 October 2011 - 05:09 PM
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however one must realize that every thing starts in a default belief and requires a burden of proof for the default belief to be abandoned. it would not make sense for believing in positive statements' validity so the only remaining is to not believe a positive statement until
- Posts: 660 | Joined: 01-July 10
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#34 24 October 2011 - 06:12 PM
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#36 13 November 2011 - 05:04 AM
The Crow, on 25 October 2011 - 11:44 AM, said:
sounds safe and like something a home chemist would want to handel
Quote
however one must realize that every thing starts in a default belief and requires a burden of proof for the default belief to be abandoned. it would not make sense for believing in positive statements' validity so the only remaining is to not believe a positive statement until
- Posts: 660 | Joined: 01-July 10
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#37 31 December 2011 - 05:54 AM
I am reluctant to go into detail as I havent had the chance (mercury is hard to come by) to test it and don't want anyone hurting or poisoning themselves trying to copy it.
So just basically, the THEORY behind the method is to:
- Form the sodium in the mercury as per the mercury cell process for sodium hydroxide production.
- Collect the amalgam
- Remove mercury from the amalgam by vacuum leaving behind sodium metal
The sodium would be contaminated with mercury- but seeing as im using this for reagents thats not so much of a problem as the mercury probs wouldn't react.
- Posts: 153 | Joined: 30-December 11
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#39 7 February 2012 - 10:50 PM

In above picture, 6.12 g KOH (flakes), 3.12 g Mg, 50 ml Shellsol D70, and 1.02 g 2-methyl-2-butanol are being heated in a sand bath to 200 °C, such a reaction (conducted in the absence of oxygen) can produce globules of metallic potassium after 2 hours.
I have been thinking about the possibility of reducing sodium hydroxide with aluminum foil. So a calculation of the expected enthalpy of formation of such a reaction may be helpful, to get some idea as to whether such a reaction would be expected to be favorable.
The enthalpy of formation for Al2O3 is -1669.8 kJ/mol, while the value for Na2O is -414.2 kJ/mol.
As Al2O3 contains 3 times as many oxygen atoms per mol, 3 times 414.2 equals 1242.6, which is still less than 1669.8, so aluminum has more affinity for oxygen than sodium. And indeed an exothermic thermite reaction between sodium hydroxide and aluminum powder can produce sodium.
http://www.youtube.com/watch?v=908rjHQ5mmc
The enthalpy of formation for AlCl3 is -705.63 kJ/mol, while the value for NaCl is -411.12 kJ/mol.
As 3 times 411.12 equals 1233.36, sodium has more affinity for chlorine than aluminum. And indeed, the reduction of aluminum chloride by elemental sodium was first done by H. Sainte-Claire Deville, although H. C. Ørsted had previously used potassium instead.
But of course the interaction with the alcohol would affect the enthalpy of formation, increasing the affinity of sodium for oxygen. A quick estimation of this effect can be made by comparing the enthalpy of formation for sodium hydroxide, which is no doubt even more favorable than sodium alkoxides (sodium alkoxides vigorously hydrolyse with water).
NaOH -425.93 kJ/mol
H2O -285.83 kJ/mol
Na2O is -414.2 kJ/mol
So the hydration of sodium oxide to anhydrous sodium hydroxide should release 151.83 kJ for each mole of Na2O reacted.
Na2O + H2O --> 2 NaOH
So it can be inferred that the presence of tert-butanol would not significantly affect the affinity of sodium for oxygen, meaning that the reduction of a sodium alkoxide by aluminum should still be energetically favorable.
The competing affinities between sodium and aluminum for fluorine apparently is more complicated:
Quote
Metallurgical and Chemical engineering, Volume 11, p178 (1913)
This post has been edited by Anders Hoveland: 7 February 2012 - 11:00 PM
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