trinitrotoluene
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Perperation of calcium
The first method is the electroylsis of CaCl2, and KCl is used to lower the melting point.
But second method is a reduction method where Al powder is used to reduce its salt into pure calcium.
2 Al+3CaCl2=3Ca+2AlCl3
Aluminium chloride vaporizes at 180*C so I guess it can be boiled off leaveing the Ca behine.
But what will be required to start the reaction? Flame? Magnesium ribbon?
TNT
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Mumbles
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I don't know if that reaction would work. I think that you might need to use CaO. When the O is driven off it could be used to fuel the
aluminum. I think an Oxidiser is a must in those reactions.
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Polverone
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The CaCl2 (or CaO) *is* the oxidizer. The calcium compound is reduced to the free metal, simultaneously oxidizing the aluminum to the corresponding
compound (Al2O3 or AlCl3).
Metallic calcium is more electropositive than metallic aluminum, so I'm guessing that it's the higher lattice energy of the aluminum
compounds that would drive this reaction.
Anybody care to confirm/deny my supposition?
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a_bab
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Well, since C or even Fe are able to replace K or Na or Cs from their compounds, Al will obviously replace Ca. It's a matter ot temperature. To
obtain K with iron around 1300 degrees C are needed.
I am researching now several methods to obtain Ca, Ba, Sr and for the Ca I guess I'll use the Moissan method one which will render pure Ca
crystals. It's based on two facts: Ca is soluble in molten Na and Ca doesn't react with ethanol.
So, 600 grams of coarsely crushed anhydrous CaI2 are placed along with 250 grams of Na lumps (as large as a nut) in an iron crucible of 1 litre
capacity. The crucible is closed with a screw lid and heated about an hour at a dull red heat then alowed to cool. The calcium which is soluble in
excess of Na at red heat separates at the point of solidification and becomes practically insoluble. The metallic portion of the melt is cut-up into
medium sized fragments and gradually introduced into absolute alcohol. The Na dissolves, leaving the Ca into brilliant white crystals (yummy !), 98.9
to 99.2 pc pure.
Well, I'll try the method along with several very exciting projcts when I'll have enough time and money and patience from my wife. The
first project will be a "mini sodium factory" though...
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Nick F
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The reaction of CaCl2 and Al will work due to the fact that AlCl3 is so volatile. It's an equilibrium reaction between Al + CaCl2 and AlCl3 + Ca,
and the vapouristaion of AlCl3 forces the position of equilibrium towards Ca, giving a high yield. Lattice enthalpy probably plays a role too
Polverone, since Al3+ is much smaller and more charged than Ca2+. But I think volatility of AlCl3 is the most important factor.
The same applies when using C or Fe as the reducing agent with metal oxides; CO2 escapes, leaving the metal, or with Fe, the metal is boiled off,
leaving iron oxides.
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Polverone
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ah, true
It would be interesting, then, to compare aluminothermic reduction of CaO vs. CaCl2, as Al2O3 is decidedly non-volatile.
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a123x
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Anyone happen to know what type of temperatur is necessary to get Al to replace Ca+ in CaCl2? I came up with a rather simple set up to do this
reaction with. Pretty much would just be a large L(except with the L rotated 90 degrees clockwise) shaped pipe with a valve located somewhere on the
long part of the L. The CaCl2 + Al would be in the shorter part with the heat source under it, a vacuum cleaner would be attached to the open end of
the long leg of the L to suck out forming AlCl3(g). When I feel the process is complete I would remove the heat and close the valve then let it cool
down then remove my Ca.
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Theoretic
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Industrially, the Ca metal obtained by electrolysis is purified from salt impurities by passing (molten?) it over aluminium shavings. So I think it
should work.
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kryss
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From Thermodynamics, data from
http://webbook.nist.gov/chemistry/
2Al +3CaCL2 -> 2AlCL3 + 3Ca
delta H = delta Hf(products-reactants)
=2x -706 - 3*(-796)
= -1412 +2384
= +1026 kj/Mol ie reaction strongly endothermic
2Al +3CaCL2 -> 2AlCL3 + 3Ca
At temp where AlCl3 is in gas phase:
S= 56+ 415 -> 628 + 123
delta S = 751-471= 280 j/molK ie
dG=dH-tdS where delta G=0
T= -1,026,000/ -280
T approx 3000 Kelvin for reaction to commence!
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Theoretic
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Apparently NOT. If this reaction is used as a purification reaction, then it must be quite easy and simple to perform, and apparently at the
temperature of the electrolysis bath (the temperature of calcium coming out of the electrolyser).
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ipr
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I had tried to melt Al with CaCl2 in air some time ago. There appear point flashes; I think these were Ca oxidizing in air. However no evidence of
vaporing AlCl3 was observed.
I have found using phase diagram CaCl2-AlCl3 that the complex salt apparently arises preventing the evaporation of AlCl3. Duplex fluid is indicated
only at high AlCl3 content. Maybe it is why no traces of reaction are observed?
There is a short reference to practical implementation of this reaction in the book by H.Remy. The reference is: Hackspill, 1951. However nothing in
the kind was mentioned in Hackspill's Chimie Minerale.
Is there someone who knows Hackspill's original papers on preparation of calcium from CaCl2? Write me please.
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unionised
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A couple of thoughts.
Why doesn't calcium react with alcohol? Mg does.
If the purification of calcium with aluminium is to remove other materials than the chloride then the reaction delta G would be different and the
reation temperature might well be a lot lower.
I suspect the removal of oxide might be easier than chloride because you gain the lattice energy of the alumina (granted, you lose that of the CaO).
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