Sigh… another wild goose chase by those unwilling to consider condition 1: Is It Possible, Thermodynamically?
CaO HoF = - 635 kJ/mol
Al2O3 = - 1676 kJ/mol
CaO + 2/3 Al --- > Ca + 1/3 Al2O3, Heat of Reaction = + 635 – 1/3 1676 = + 76 kJ/mol, reaction is (strongly) endothermic and
unless you have a trick up your sleeve to force the equilibrium from left to right nothing will happen.
Calcium is in fact occasionally used where aluminium fails as an oxide reducing agent, the technique is then called calciothermy. V2O5 has been
reduced with Ca for instance and early production of metallic U (from UO2) also used Ca (now Mg with UF4). Drawback of Ca is that CaO has an insanely
high MP, not desirable in most oxide reductions.
One potential trick is to heat a mitxure of CaO and Al to very high temperature under vacuum: Ca has a lower BP (1,484C) than Al (2,519C) and the Ca
could thus in (unsubstantiated) theory be selectively boiled off. Good luck with that!
Ca is produced electrochemically because it’s the only PRACTICAL way.
One could of course envisage various reducing agents in very extreme conditions and never be able to bench test it because the conditions required.
CaO + C --- > Ca + CO is a possibility because at very high temperature (x,000 C) the volatile Ca and CO drive the equilibrium to the right, IF
you can successfully remove them form the mix. That would involve probably high vacuum on top of the high temperatures. Again, bonne chance!
The communications published above are likely by bullshitting freewheelers…
[Edited on 16-12-2010 by blogfast25] |