Sciencemadness Discussion Board

Black Aluminum

Killer - 1-4-2013 at 15:41

Hello everyone. I have already made many small batches of the stuff, but was just curious... Is there any way to make it faster? Tumbling for days is ridiculous. I wouldn't mind if I could use a 55 gallon drum ( any dangers there?) and just get it done and over with, but to use a tumbler and get an almost insignificant amount just a pain in the ass. I mean if that is the way it has to be I can deal with that, but it just seems like there has to be a faster or larger way to do it. Did some searches on it here and didn't come up with what I was looking for. Someone suggested just running it in a blender for a few hours but I thought that would be dangerous and I don't quite see the tumbling effect being completed... Any help would be sweet.

plante1999 - 1-4-2013 at 16:11

Your best way is a ball mill, else than that you will have to make a expensive machine that make atomized aluminum.

Killer - 1-4-2013 at 16:25

Yea that is what I use but it is small. Just curious, do you think a big plastic container to tumble in would be better than metal? The small tumbler seems to be pretty safe, curious about using larger amounts... IE. Making a larger tumbler. Just thinking some unforeseen problem will arise in doing so. Any comments or suggestions appreciated. Well after days of reading and finally finding my wy here, I guess a dryer motor with a couple pulleys and a five gallon bucket ought to do it. Thanks again.

[Edited on 2-4-2013 by Killer]

Fantasma4500 - 1-4-2013 at 22:11

i think you should try melting it together with some magnesium, first aluminium tho, and keep oxygen and whatever not away from the melting pot when adding magnesium, seal it pretty well, i believe magnesium can even react with nitrogen!

if you didnt know already the idea of magnalium is that its actually really easy to crush to a powder thats very fine, a person i know have crushed up ~200 mesh into +800 mesh within a few days of ballmilling..
burnrate fits the meshes (:

edit:
add charcoal powder of any sort to the magnalium to keep it from oxidizing and suddenly burst into flames, which can as dustcloud be somewhat violent, and most of all very hot, charcoal isnt to act as fuel but just to cover the magnalium up abit..


[Edited on 2-4-2013 by Antiswat]

AJKOER - 4-4-2013 at 08:35

OK, here is a idea (somewhat speculative at this point), and it does involves absolute Ethyl alcohol, C2H5OH (perhaps adding K2CO3 to less than anhydrous C2H5OH and salting-out the water may work to a sufficient degree, or not). Obviously, try this experiment on a small scale first (no need to waste good booze).

First, burn some Al in Chlorine forming a small amount of anhydrous AlCl3 (but if you don't have dry C2H5OH to start with, you might as well use the hydrated Aluminum chloride). Then add the anhydrous and heated C2H5OH to the AlCl3 and proceed to slowly dissolve the pretreated Aluminum (with NaOH) with constant stirring (which is apparently important, for more details, see full reference paper at http://www.jim.or.jp/journal/e/pdf3/50/06/1433.pdf ).

Next, insert large strips of Mg to displace the Al. Scrap off and appropriately store.

Chemistry:

2 Al + 6 Et-OH --> 2 Al(O-Et)3 + 3 H2 (g)

and, more speculation:

3 Mg + 2 Al(O-Et)3 --> 3 Mg(O-Et)2 + 2 Al (s)

Note, in the current context on the dissolving of the Al in the presence of AlCl3, the alkoxide reaction on the Al can be described by the following electrochemical reactions:

Al = [Al]3+ + 3 e as the anodic reaction.

and

2Et-OH + 2 e = 2 [Et-O]- + H2 as the cathodic reaction.

Note, normally, in an aqueous environment:

2 Al + 6 H2O --> 2 Al(OH)3 + 3 H2 (g)

but here, there is limited to no water.

With unlimited time, any Al formed could redissolve recreating Al(O-Et)3 and liberating Hydrogen, but with sufficient excess Magnesium and scrapping to recover surface exposure, more Mg(O-Et)2 and Al is formed with the net effect that all the excess C2H5OH is consumed.

Also, no matter how slowly the first reaction of dissolving Aluminum is (say, from the presence of water), once it has begun, slowly water is removed, and the process speeds up due to the hydrolysis of the Al(O-Et)3, which removes water forming a precipitate of Aluminum hydroxide and reforming Ethyl alcohol:

Al(O-Et)3 + 3 H2O --> Al(OH)3 (s) + 3 Et-OH

[EDIT] Potential issues with this synthesis: drying the Ethanol to the proper level (?) as it is primarily an electrochemical process, and will the Mg substitution reaction actually proceed in this medium possibly as an electrochemical reaction also?

[EDIT][EDIT] Speculation: There is a potential that the electrochemical reaction may be significantly improved (and accelerated) with the negative impact of water resolved (by its electrochemical decomposition) upon the application of an external current. The Aluminum to be dissolved (and ultimately converted to a fine reactive form) will serve as the anode and a silver wire could serve as the cathode. The anode would be replaced with Magnesium to produce the final Aluminum product.

[Edited on 4-4-2013 by AJKOER]

[Edited on 4-4-2013 by AJKOER]

Metacelsus - 4-4-2013 at 11:41

Aluminum alkoxides SHOULD be reduced by Mg, but the reaction might either aluminum-plate the Mg (ending the reaction) or proceed at a slow rate. Also, finding a solvent that doesn't react with either Al or its alkoxides might be tricky.

Just ball mill it.

AJKOER - 4-4-2013 at 15:04

Quote: Originally posted by Cheddite Cheese  
Aluminum alkoxides SHOULD be reduced by Mg, but the reaction might either aluminum-plate the Mg (ending the reaction) or proceed at a slow rate. Also, finding a solvent that doesn't react with either Al or its alkoxides might be tricky.

Just ball mill it.


Yes, I agree, but I believe the thread question is alternatives to mechanical means.

Now, I recalled and also found a comment by one of our own (Arrhenius at http://www.sciencemadness.org/talk/viewthread.php?tid=12831 ) who stated that "I2 is a fine way of initiating the reaction of alcohols with magnesium or aluminum".

Now, I suspect this is true even in the presence of water by the following reactions:

2 Al + 3 I2 --> 2 AlI3

but, in the presence of water, a hydrolysis occurs:

2 AlI3 + 6 H2O --> 2 Al(OH)3 + 6 HI (see http://en.wikipedia.org/wiki/Aluminium_iodide )

removing some water. Also, with the newly created HI:

2 Al + 6 HI --> 2 AlI3 + 3 H2 (see Wiki same link)

or, for all three reactions the net reaction is:

4 Al + 6 H2O + 3 I2 ----> 2 Al(OH)3 (s) + 3 H2 (g) + 2 AlI3

and, upon adding 6 more H2O to each side:

4 Al + 12 H2O + 3 I2 ---> 4 Al(OH)3 (s) + 3 H2 (g) + 6 HI

and adding 2 more Al to each side:

6 Al + 12 H2O + 3 I2 ---> 4 Al(OH)3 (s) + 6 H2 (g) + 2 AlI3

Note, the process can be repeated but the quantities 3 I2 and 2 AlI3 will remain constant while the amount of Al and H2O increase. As such, the Iodine acts essentially as a catalyst (only a small amount is required to initiate the reaction regardless of the amount of water, and only that initial Iodine is consumed). This analysis can also suggest the amount of excess Aluminum to employ to remove the unwanted water from the alcohol (my calculations are 2/3 moles Al are needed to be added for each estimated moles of H2O to be removed). Also, note that when all the water is consumed, the chain breaks and the initial number of moles of Iodine added is now 2/3 as many moles of AlI3. Also, this dehydration reaction generates gas (Hydrogen), so a closed reaction vessel is not recommended.
--------------------------------------------------------------

Now, on the plating issue, if it can easily be scrapped off, there is no problem, else, I agree, a problem to be addressed.


[Edited on 5-4-2013 by AJKOER]