Mechanism of conversion of Cr from FeCr2O4 (chromite) The basic steps involved in the conversion of $\ce{FeCr2O4 -&gt; Cr}$ involves,

$$\ce{FeCr2O4 -&gt;[NaOH / air, \Delta] Na_2CrO4 -&gt;[C, \Delta] CrO3 -&gt;[Al, \Delta] Cr}$$

Basically, I'm having problems in learning so large amount of inorganic reactions, so I thought it would be easy to learn the mechanism behind these reactions, just like we do in organic chemistry.
Therefore, I'm keen to know the mechanism behind first conversion (i.e., from $\ce{FeCr2O4 -&gt; Na_2CrO4}$). I've already tried to search for this, but found nothing related.

My attempt
I know some basic sets of inorganic mechanisms, as follows

Ion-exchange
Decomposition
Water-addition
Redox
Complex reaction

From the above listed, I don't think it would be ion-exchange (bcoz, $\ce{FeCr2O4 -&gt; FeCr+ + CrO4-}$ is very unlikely to happen), water-addition or complex reaction. I guess it could be decomposition or redox, but I'm not sure on the &quot;bond-movement&quot; mechanism part.
The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever had to state. It is :    $$\ce{FeCr_2O_4 + 11 OH^- -&gt; Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write.
$$\ce{O_2 + 2H_2O + 4 e-  -&gt; 4 OH^-}$$ The overall equation is obtained by multiplying the first equation by $4$ and the second by $7$ and adding the whole. After simplification, it yields :
$$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -&gt; 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$
or, without ions :
$$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -&gt; 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$
This was the first step, defining how to pass from the mineral $\ce{FeCr_2O_4}$ to $\ce{Na_2CrO_4}$, at high temperature, and in the presence of air and $\ce{NaOH}$. The final mixture can be washed with water, which dissolves easily $\ce{Na_2CrO_4}$, as $\ce{Fe(OH)_3}$ is insoluble in water.
Then $\ce{Na_2CrO_4}$ is transformed in $\ce{CrO_3}$ by adding moderately concentrated sulfuric acid : $$\ce{Na_2CrO_4 + H_2SO_4 -&gt; CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ is not soluble and can be separated by filtration. Then it is mixed with aluminum powder to get a exothermic reaction when engaged by a match : $$\ce{2 Al + CrO_3 -&gt; Al2O3 + Cr}$$