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Author: Subject: Is this for Real? Mg + NaOH ==> Sodium?
aliced25
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shocked.gif posted on 22-5-2013 at 22:14
Is this for Real? Mg + NaOH ==> Sodium?


I watched this video on youtube and I just had to wonder, is this really sodium (what he drops on the water looks like it)?

http://www.youtube.com/watch?v=seSg_GWj1b0

If so there are several ways of improving yields - mineral oil and then heat to melt the sodium and decant it off the slag/unreacted Mg/NaOH/etc.




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[*] posted on 22-5-2013 at 22:41


Yes it is
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[*] posted on 22-5-2013 at 23:09


Can you elaborate on your ideas to improve the yield?



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[*] posted on 23-5-2013 at 00:32


Shit, where to start - fused sodium hydroxide - out of the bottle it will contain water to some degree which will destroy yield. Inert atmosphere, even oxygen free atmosphere to stop oxidation (or get some argon and pump it in on top), rinse the magnesium powder as it is usually coated (thus reducing yield burning it off) and then adding in sufficient mineral oil/paraffin to allow for separation of the melted metal from the slag without water. Other people will have even better ideas, but those are some. You could also look into adding the mineral oil and melting the sodium (around 97C), then collect the balls of sodium that coalesce in the paraffin.

FUCK ME - CAVEMAN CHEMISTRY TO SODIUM

[Edited on 23-5-2013 by aliced25]




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[*] posted on 23-5-2013 at 03:57


You have read the potassium thread, yes?



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[*] posted on 23-5-2013 at 05:27


Forget about 'significantly improving yield', it's been tried HERE before (search for it before you waste more valuable time). This NaOH + Mg 'thermite' is a messy reaction, which works only to produce miniscule amounts of very impure sodium and for demo purposes. It is NOT a practical way to produce sodium of any appreciable quality in any appreciable quantity.

Sorry to rain on anyone's parade. :(

[Edited on 23-5-2013 by blogfast25]




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[*] posted on 23-5-2013 at 09:22


Quote: Originally posted by blogfast25  
Forget about 'significantly improving yield', it's been tried HERE before (search for it before you waste more valuable time). This NaOH + Mg 'thermite' is a messy reaction, which works only to produce miniscule amounts of very impure sodium and for demo purposes. It is NOT a practical way to produce sodium of any appreciable quality in any appreciable quantity.

Sorry to rain on anyone's parade. :(


That's what I was thinking, I was hoping the OP had some novel way to do this better. I have tried many times but it's not worth it like you say. WAY too contaminated and small amount.




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[*] posted on 23-5-2013 at 10:06


Quote: Originally posted by chemcam  
That's what I was thinking, I was hoping the OP had some novel way to do this better. I have tried many times but it's not worth it like you say. WAY too contaminated and small amount.


Not sure what you mean by OP.

The only magnesiothermic and aluminothermic reactions that yield good metal are those that fulfil the following conditions:

1) Must generate enough heat so both the metal and slag (MgO or Al2O3 respectively) form in the liquid phase, forming a melt that allows metal and slag to separate out.

2) End temperature of the reaction must not exceed the boiling point of the metal produced, as otherwise it will boil off (often leading to potentially dangerous flashing).

Reactions involving Al or Mg and NaOH, KOH or LiOH (never mind the other alkali metal hydroxides) do NOT satisfy these conditions and lead to messy, sintered masses from which decent metal recovery is basically impossible. A few small nuggets, yes. Anything more, no.




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[*] posted on 23-5-2013 at 14:50


2) can be solved partially by performing the reaction in a bomb. I believe one member had ignited such a thermite mixture in a suitable steel bomb, and was rewarded with a buttery wad of metal.

Bombs suitable for high pressure and extremely high (thermite-grade) temperatures are a rather dubious proposition, however...

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[*] posted on 23-5-2013 at 15:02


Not talking about anything quite so crude - but there is fuck all on the web about magnesiothermic reduction of sodium, which is what got me excited.

A propane furnace, a mild-steel pipe (low-carbon) and a boat - then a steel water-cooled condenser and a collection cup. The vacuum needed is fairly high compared to what people expect (according to the papers) and it would give high purity sodium.

One interesting question is whether vacuum would be needed at all... Magnesium burns at 3,100C (auto-ignition in air of 473C - http://en.wikipedia.org/wiki/Magnesium) which is well above the boiling point of sodium (883C - http://en.wikipedia.org/wiki/Sodium), so even a moderate vacuum - or flow of argon should push the vaporised Na onto the condenser (I'd personally go for the moderate vacuum), allowing for the collection of Na without the slag (see the 2nd & 3rd papers, one has a design of a pipe in a v-shape to collect Li the same way). If it worked, it could also be applied to Lithium (BP 1,342C -http://en.wikipedia.org/wiki/Lithium).

It all depends on how Magnesium burns in a modest vacuum (say 1/2 - 1/4 atm). Or use CO2 as a sweep gas, Mg burns in iit, and it won't affect the metals (or their hydroxides/oxides) in the absence of water.



Attachment: Kroll.Schlecten.Laboratory.Preparation.of.Lithium.Metal.by.Vacuum.Metallurgy.pdf (373kB)
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Attachment: Epstein.Howland.The.Distillation.of.Lithium.Metal.pdf (506kB)
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Attachment: Rogers.Viens.Effect.of.Pressure.on.the.Refining.of.Lithium.by.Distillation.pdf (465kB)
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Attachment: Rutkauskas.Determination.of.the.Oxygen.in.Sodium.by.Vacuum.Distillation.pdf (910kB)
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Attachment: Loshkin.Manufacture.of.High.Purity.Sodium.by.Distillation.pdf (57kB)
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[Edited on 23-5-2013 by aliced25]




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[*] posted on 24-5-2013 at 09:32


Quote: Originally posted by aliced25  
Not talking about anything quite so crude - but there is fuck all on the web about magnesiothermic reduction of sodium, which is what got me excited.



There is 'f*ck all' on the web about this because it just isn't very promising at all. This is one of many SM's projects that probably won't even come off the ground or even the drawing boards.

Elsewhere a 'Zan Divine' posted some amazing stuff on reducing CsCl with Li in vacuo, with excellent results and fairly simple apparatus. I suggest you have a look at that re. reduction of NaCl with Li.

Merely for comparative purposes: Li, BP = 1342 C; Na, BP = 883 C; Cs, BP = 670 C; LiCl, BP = 1382 C.

Reduction of NaCl with Mg in vacuo could also be considered because the BP of MgCl2 is 1412 C. With a melting point of Mg at 650 C (BP = 1091 C), the 'sweet spot' mix of volatilities may allow [note caveat!] Na to be distilled off from a mixture of NaCl and Mg at > 700 C.



[Edited on 24-5-2013 by blogfast25]




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[*] posted on 24-5-2013 at 09:44


Quote: Originally posted by 12AX7  
2) can be solved partially by performing the reaction in a bomb. I believe one member had ignited such a thermite mixture in a suitable steel bomb, and was rewarded with a buttery wad of metal.


Tim


I think you are remembering wrongly but I'll gladly acknowledge my mistake if you can dig it up.

In a bomb you have the additional problem that sodium hydride could be formed.

With a Standard Enthalpy of Formation of - 56 kJ/mol for NaH, the reaction:

NaOH + Mg === > NaH + MgO is thermodynamically favourable to:

NaOH + Mg === > Na + MgO + 1/2 H2

In open crucible conditions the second one is favoured because H2 leaves the reaction, in bomb conditions it cannot do so.




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[*] posted on 24-5-2013 at 09:49


To get NaCl to react with lithium wouldn't it need to be in solution? What happens to the Na? NaOH right?
I read though a thread about this before and you said reductions of chlorides with lithium is not possible. Have to changed your mind? Or am I misunderstanding the context?

http://www.sciencemadness.org/talk/viewthread.php?tid=15061&...




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[*] posted on 24-5-2013 at 10:04


Quote: Originally posted by chemcam  
Or am I misunderstanding the context?



You are misunderstanding the context. Here we are exploiting the differences in volatitity so that, under vacuum, the desired species volatilises, thus leaves the reaction mix, thus pulling the equilibrium to one side. Take the example of:

CsCl(s) +Li(l) < === > Cs(l) + LiCl(s)

In the conditions relevant to the thread you linked to this equilibrium stays very much to the left and nothing much happens. But at high temperature and under vacuum, Cs metal is the most volatile of the four species and is distilled off. This pulls the equilibrium to the right because it reduces the concentration of the Cs in the mix ('Mass Effect').

This little 'trick' is exploited in numerous industrial processes, among others in blast furnaces where the volatile CO/CO2 reaction product's 'escape' pulls the reduction of iron oxide to the Fe side. In conditions where the carbon oxides could not leave the reaction, carbon has very little reducing capability.



[Edited on 24-5-2013 by blogfast25]




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[*] posted on 24-5-2013 at 13:49


Quote: Originally posted by chemcam  
To get NaCl to react with lithium wouldn't it need to be in solution? What happens to the Na? NaOH right?
I read though a thread about this before and you said reductions of chlorides with lithium is not possible. Have to changed your mind? Or am I misunderstanding the context?

http://www.sciencemadness.org/talk/viewthread.php?tid=15061&...


The reaction will be between a solid (NaCl) and a liquid (Li).
Because of the low pressure in the system, the liquid Na will boil off and you collect the vapors in a condenser.

The sodium won't be pure because some of the lithium boils off as well. Also, the solid NaCl will get covered by solid LiCl. There will be crystal defect because lithium ions are small compared to sodium ions.
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[*] posted on 24-5-2013 at 15:27


well you can make silicon metalloid by SiO2 + Al + S
the Al and S first reacts whereafter the SiO2 takes part in the reaction giving silicone metal..

in other words, a much slower burning thermite should be possible to do, perhaps with something to kick start it such as.. sulfur mixed in? and with aluminium
i might be wrong tho, not sure about reactivity and all of that (:




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[*] posted on 24-5-2013 at 18:34


Quote: Originally posted by blogfast25  

Merely for comparative purposes: Li, BP = 1342 C; Na, BP = 883 C; Cs, BP = 670 C; LiCl, BP = 1382 C.

Reduction of NaCl with Mg in vacuo could also be considered because the BP of MgCl2 is 1412 C. With a melting point of Mg at 650 C (BP = 1091 C), the 'sweet spot' mix of volatilities may allow [note caveat!] Na to be distilled off from a mixture of NaCl and Mg at > 700 C.


Sounds more like the Ames process (http://en.wikipedia.org/wiki/Ames_process), except using the Chloride instead of the Fluoride and Sodium instead of Uranium.

It would be an interesting procedure to work on, especially as NaCl is fairly easily acquired dry and a basic tube furnace (propane or electric) would get the temperatures high enough. Cs would be an interesting addition to the collection.




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[*] posted on 25-5-2013 at 04:21


The Ames Process is substantially different from methods that require distillation, in the sense that it relies on the very substantially exothermic reaction UF4 + 2 Mg == > U + 2 MgF2, with a Standard Heat of Reaction of about - 354 kJ/mol (of UF4). This enthalpy, combined with the preheat soak to about 600 C, ensures the reaction proceeds spontaneously (and quite quickly) and results in an end temperature well above the MPs of both MgF2 and U, which then separate out gravitationally (with U that really works a treat!). It's very much like a preheated Goldschmidt reaction.

Compare this to (for instance) CsCl + Li < === > Cs + Li, there the Standard Heat of Reaction is about + 35 kJ/mol (of CsCl), so actually endothermic. The reduction only works by removing the most volatile component (Cs) from the mix by distillation, thereby 'pulling' the equilibrium to the right by mass effect.

They really are very different separation techniques.


[Edited on 25-5-2013 by blogfast25]




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[*] posted on 25-5-2013 at 04:33


Quote: Originally posted by Antiswat  
the Al and S first reacts whereafter the SiO2 takes part in the reaction giving silicone metal..



Nope. The reactions:

SiO2 + 4/3 Al === > Si + 2/3 Al2O3

And:

2 Al + 3 S ===> Al2S3

... proceed more or less consecutively. The first one simply doesn't generate enough heat to obtain a molten mixture of Si and Al2O3. To boost heat output, the second, highly exothermic reaction is used, with the required ingredients simply mixed in with the SiO2 and Al powders. Everything then just 'burns' together. It's called 'heat boosting'. The other advantage is that Al2S3 has a much lower MP than Al2O3, thereby reducing the viscosity of the melt and promoting separation between the Si and slag.

Another widely used [in industry] heat booster system is KClO3 + 2 Al === > KCl + Al2O3 which is extremely exothermic. Crude niobium metal is produced from pyrochlore, aluminium and potassium chlorate (or sodium chlorate) to ensure the insanely high MP of Nb is reached.

Quote: Originally posted by vmelkon  
The sodium won't be pure because some of the lithium boils off as well.


I'm not convinced that the contamination will be significant if you get the conditions right. See the CsCl + Li example. Sure, Cs is more volatile than Na but Li is much less volatile than both.



[Edited on 25-5-2013 by blogfast25]




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[*] posted on 27-5-2013 at 11:56


Na boils at 890 C. K at 760 C.

What is the least active metal that can be used to reduce Na and K to metal vapour? (You need the metal to form a refractory compound, too).

Another relatively active but volatile metal is Zn. Can you similarly boil off Zn reduced by a less active but less volatile metal, like Fe, Sn or something similar?
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[*] posted on 27-5-2013 at 13:29


It's not really about 'activity'. The reducing agent must have volatility much below that of the target metal. In principle that works for Zn too.



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[*] posted on 28-5-2013 at 04:24


The standard metals used in the Pidgeon Processes are silicon, aluminium, magnesium, calcium and sodium. All of which can also be made by the Pidgeon Process as well. I'm moving this to the unconventional sodium thread if the moderators would please help?

A basic Propane kiln/furnace will provide sufficient heat to allow the reaction to proceed, while I suspect a simple sliding vane (10pa/75mtorr) would suffice for most of these processes. I'd also like to look at Phosphorus by silicothermic reduction, a metaphosphate finely ground and briquetted with finely ground silicon should give the alkali silicate and P vapour, at comparatively low temperatures.

The best routes are through the respective oxides, in the case of Lithium & Sodium for example, these come from either decomposition of the Carbonate at high-temperatures or formation and decomposition of the peroxide. I'm not sure where one would even look for Cesium Oxide/Carbonate or whatever, or why I'd want to? Is there any real advantage in having it?;) Apart from Element Collection?

There is a blast furnace design - lab scale - for P-Vapor from a preheat, note the temperatures for the Na/K Metaphosphates? Imagine using Si as the reductant and in a vacuum?



[Edited on 28-5-2013 by aliced25]




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[*] posted on 28-5-2013 at 05:03


Not sure what you mean by 'a simple sliding vane'...



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[*] posted on 28-5-2013 at 08:35


Quote: Originally posted by blogfast25  
Not sure what you mean by 'a simple sliding vane'...
I read that as "rotary vane vacuum pump". Though gas combustion kilns generally heat from the inside, which makes vacuum basically impossible. And if you were to use a vacuum muffle, you've got a sealing problem, since the seal has to be at kiln temperature. (Vacuum seals are usually not in the hot center of a kiln.) Easiest would be electrical heat, but even getting power feed-through at kiln temperatures that are proof against vacuum isn't trivial. All-in-all seems kind of half-baked to me.
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[*] posted on 28-5-2013 at 11:20


Ah, I see, a reference to the vacuum pump itself. But from the way I understand it the kiln doesn't have to be under vacuum: only the reactor placed in it.



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