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For the extraction of chromium, fusion of chromite ore with sodium or potassium carbonate in excess of air is done. What exactly happens in this 'fusion' process?

According to my textbook (class XII chemistry, NCERT): d- and f-block element, extraction of chromite ore

According to the dictionary: To melt or fuse (ores) in order to separate the metallic constituents.

Now, while this makes sense and is sort of (implied) by my textbook? how exactly is this process done? I want to learn the technicalities related to it! Further I've heard we use alkali metals in fusion, why do we use them ?

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  • $\begingroup$ Why "excess of air"? Is the iron (II) supposed to oxidise to iron(III)? You definitely need an excess of carbonate. $\endgroup$
    – Karl
    Feb 8 '20 at 13:52
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    $\begingroup$ it came up for me in my highschool textbook : ncert chemistry class 12 textbook. I asked the question because I could not find any answer to this directly in my textbook nor the internet from a direct search. $\endgroup$
    – 666User666
    Feb 13 '20 at 2:52
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    $\begingroup$ Very interesting that you are placing multiple 500 bounties, on multiple stack exchanges, not just this one. Respect! ;-) $\endgroup$
    – Ed V
    Sep 26 at 12:52
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    $\begingroup$ I had more than thousand but I gave away both. As soon as I started the bounty, it was substracted from my account @JamesGaidis $\endgroup$
    – 666User666
    Sep 26 at 14:26
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    $\begingroup$ @Buraian: Thanks. Very generous of you! Keep earning! $\endgroup$ Sep 27 at 12:59
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"To fuse" is another word for "to melt" (e.g. "heat of fusion"). Specifically, if you say you want to fuse two materials, you melt them in the hope that they will mix.

In this case, you melt the carbonate, and hope that the chromite will dissolve in it. Because e.g. $\ce{Cr2O3}$ has a melting point of $\pu{2435 °C}$, chromite ($\ce{Fe(II)Cr2O4}$) typically around $\pu{2140 °C}$. That is out of reach with a common Bunsen burner, and even if you did reach it, it would still be the same insoluble mess after cooling down.

When your chromite is dissolved in alkali carbonate, the mixture easily dissolves in $\ce{HCl}$, and you can properly run your wet analytics.

Soldering for example is a very similar process: Your soldering iron cannot nearly melt copper, but the molten tin/lead/etc. fuses (partly dissolves/melts) with the copper surface, and you get a continous metallic connection.

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  • $\begingroup$ RE: "if you say you want to fuse two materials, you melt them" -- You're really only melting the alkali carbonate which then chemically reacts with the ore. $\endgroup$
    – MaxW
    Feb 9 '20 at 14:30
  • $\begingroup$ @MaxW I didn´t want to split too many hairs in the first two sentences. ;) $\endgroup$
    – Karl
    Feb 9 '20 at 14:47
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As a comment of Karl's message, I would like to add that chromite $\ce{FeCr_2O_4}$ is the most important Chromium mineral. When mixed with $8$ times its weight of sodium carbonate, and heated to high temperature, $\ce{Na_2CO_3}$ melts at $850°$C and reacts with chromite and air according to

$\ce{8 Na_2CO_3 + 4 FeCr_2O_4 + 7 O_2 -> 8 Na_2CrO_4 + 2 Fe_2O_3 + 8 CO_2 }$.

In this equation, both $\ce{Fe}$ and $\ce{Cr}$ atoms are oxidized. And this produces a mixture of soluble sodium chromate and insoluble iron oxide, which is separated from the chromate by dissolution into water.

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    $\begingroup$ I find this oxidation scheme a bit suspect. Have you got a source for it? Where would the oxygen come from? $\endgroup$
    – Karl
    Feb 8 '20 at 19:58
  • $\begingroup$ The oxygen would be from the "excess of air" as mentioned in the OP $\endgroup$
    – Waylander
    Feb 8 '20 at 20:18
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    $\begingroup$ @Waylander How do I add an "excess of air" to a molten blob of carbonate? $\endgroup$
    – Karl
    Feb 9 '20 at 8:52
  • $\begingroup$ Bubble compressed air through it with a needle $\endgroup$
    – Waylander
    Feb 9 '20 at 9:04
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    $\begingroup$ In the industry, they add sodium or potassium nitrate to $Na_2CO_3$ to produce the necessary Oxygen. The reaction is : $$26 Na_2CO3 + 20 FeCr_2O_4 + 28 NaNO_3 --> 40 Na_2CrO_4 + 10 Fe_2O_3 + 26 CO_2 + 14 N_2$$ $\endgroup$
    – Maurice
    Feb 9 '20 at 16:28

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