Mechanism of conversion of Cr from FeCr2O4 (chromite)
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 : $$\ce{2 Al + CrO_3 -&gt; Al2O3 + Cr}$$