Sciencemadness Discussion Board

Seperating calcium and magnesium

12AX7 - 20-12-2005 at 13:44

(With some good results I may post this under the prepub forum.)

If you live in an area with dolostone bedrock, you've got a whoooole lot of calcium, magnesium and carbon around you. The problem is, depending on what you want to do with it, you need to burn a lot of reagent to use it.

I'm not interested in CO2 so I'll just let that go to waste. If you want to store it somehow (perhaps with a refrigeration compressor), you're welcome to try.

I'm more interested in seperating the lime, which can be used for synthesis, from the magnesia, which can be used for refractory. (Dolomite alone can be used for refractory, but it isn't as stable.)

Anyways, the basic idea is, you start with a rock that used to be just limestone, but was altered to contain some amount of magnesium (usually on the order of equimolar amounts, i.e., CaMg(CO3)2, though it is quite variable). Dolomite (the component mineral) is reasonably stable, so you can't just dissolve it (though that does work over time, hence caves with deposits of calcite forming inside dolomite rock). You can add acid to dissolve both, but to recover the alkali you either need to decompose the anion (cough!) or add a stronger base, which in some cases defeats the purpose entirely!

Grinding to a powder isn't going to help much, since it's a crystal lattice. You could leach the magnesium ions off the surface, but then you need to bite through calcium ions before you can get to more magnesium.

I propose destroying the structure altogether, by calcination. Heating to a yellow heat (1000C or 1800°F or so) causes loss of the CO2 content, leaving CaO and MgO loosely stuck together.

If you fire to a higher temperature (perhaps white heat, circa 1500C/2500F), you can get minor recrystallization and probably better seperation (CaO and MgO do not form a compound), though you also get lower reactivity.

This can be powdered and leached with solvents or treated with reagents. For example, hydrated lime is slightly soluble in water and could be leached, in the absence of CO2 (a 5 gallon pail sized Gregar extractor plus a hotplate and a few weeks might do it). Alternately, you could add a salt such as magnesium sulfate or chloride to precipitate insoluble magnesium hydroxide, leaving the calcium salt in solution: MgX + Ca(OH)2 --> Mg(OH)2 + CaX (X = Cl2, SO4, NO3, etc.).

I like the possibility of a large Gregar extractor, because you'll be able to grow large crystals of Ca(OH)2, and that would be cool to have on hand, if hard to stabilize!

Once you have the materials seperated, you can do whatever you want with them. CaCl2 can be used for calcium ions, such as for precipitating sulfate ions from solution (if not as powerfully so as say, barium), or dried (at expense of hydrolysis) for dessicant. Both hydroxides can be dehydrated; CaO can be used for dessicant and cement type uses, while MgO and Mg(OH)2 can be used for cements (such as phosphate cements) or a variety of ceramic and refractory uses.

Tim
(Ya, rambling on, like I do when I'm tired...)

vulture - 20-12-2005 at 14:40

Don't you think it's more economical to buy lime or CaCO3?

neutrino - 20-12-2005 at 14:47

I doubt this would be worth it. For the cost of running that plate for that long or using that much kiln fuel, you could easily buy a large amount of limestone or some other form of lime.
Anything we can do, industry does a hundred times faster and a thousand times cheaper. This is one of those things like making your own glass tubing, it just isn't worth it.

12AX7 - 20-12-2005 at 19:10

Yeah, I already have limestone...it's mined locally...except it's composed of dolomite, not calcite. See? :P

Fuel doesn't bother me. Just toss some rocks in with the usual kiln load, bisque fire and you've got some lime, BFD.

Economy has nothing to do with it, besides. Surely you know this by now...

Tim

Nerro - 21-12-2005 at 11:52

If you don't mind going through the trouble you could of course melt the mix and electrolyse with 2,38V < U < 2,76V to obtain Mg and thus be left with Ca at the cathode :D

[Edited on Wed/Dec/2005 by Nerro]

neutrino - 21-12-2005 at 14:55

Theoretically, that might work. In practice, though it wouldn't be so easy due to overvoltage, varying of redox potentials at nonstandard conditions...

12AX7 - 21-12-2005 at 15:30

At that temperature, you might as well include some carbon and use density seperation to condense the Ca/Mg vapor seperate from the CO, which can be recycled through other processes.

(Read about lunar material processing, fun stuff when you have abundant solar energy and a hard vacuum :P )

Tim

JohnWW - 23-1-2006 at 14:17

Of course, electrolysis of the molten salts, usually the chlorides or fluorides, at different voltages would be the way to separate out the Ca and Mg as metals, if they are wanted in metallic form. The calcium electrolysis would have to be done in the absence of oxygen. The gas evolved at the anode, being a highly poisonous and corrosive halogen, would have to be carefully collected, and bottled for use as a chemical byproduct.

If they are wanted as simple compounds, noting that reduction to the metals requires enormous amounts of energy, one must find a pair of their compounds with greatly differing solubilities, for a simple aqueous precipitation separation. The most promising candidates for this are the sulfates, MgSO4 being MUCH more soluble than CaSO4, alhough even then the separation is only of "technical" or "reagent" quality, not complete. The hydroxides are both only slightly soluble in water. (See Perry chapter 3 and the HBCP). A better although more costly method would probably be passing a mixture of the chlorides in solution through an ion-exchange resin column of some sort, the resin specifically absorbing one or the other.

BTW natural calcite and dolomite also contain small amounts of strontium, with its properties and those of its compounds being much closer to those of Ca than Mg. Impure limestones located near areas of oxidized minerals containing heavy metals may also contain varying amounts of Fe(II), Co(II), Zn(II), Mn(II), Cd(II), Cu(II) etc. as carbonates.

12AX7 - 24-1-2006 at 14:16

Quote:
Originally posted by JohnWW
If they are wanted as simple compounds, noting that reduction to the metals requires enormous amounts of energy, one must find a pair of their compounds with greatly differing solubilities, for a simple aqueous precipitation separation. The most promising candidates for this are the sulfates, MgSO4 being MUCH more soluble than CaSO4


Well yeah, that's not a problem, but running e.g. a five gallon Gregar Extractor would probably be cheaper than buying the sulfuric acid, wasting it, then buying and wasting the soda to neutralize it back to seperate hydroxides. Else the sulfate can be recovered, with much work and even more heat, but...pffbt... that's worse than buying acid and base.

I've done a small scale of your proposed solution BTW. I assayed the rock as 1:1 Ca to Mg (+/-10%), IOW, dolomite is the main mineral present.

Tim