Chemistry - Mechanism of conversion of Cr from FeCr2O4 (chromite)

Solution 1:

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 : $$\ce{FeCr_2O_4 + 11 OH^- -> Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write. $$\ce{O_2 + 2H_2O + 4 e- -> 4 OH^-}$$ 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 : $$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -> 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$ or, without ions : $$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -> 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$ 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 : $$\ce{Na_2CrO_4 + H_2SO_4 -> CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ 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 : $$\ce{2 Al + CrO_3 -> Al2O3 + Cr}$$

Solution 2:

This is an example of industrial recovery process of $\ce{Cr}$ from toxic chromium(VI) waste, which usually use high temperature reactions. However, the other answer is acceptable since OP's main focus is on the first reaction of converting $\ce{FeCr2O4}$ (one of the products from initial reaction of chromium(VI) waste) to $\ce{Na2CrO4}$. Because these reactions are done in high temperatures (usually, $\pu{1000-1200 ^\circ C}$; Ref.1), I'd like to make few comments:

1. $\ce{CrO3}$ would decompose to $\ce{Cr2O3}$ upon heating above $\pu{197 ^\circ C}$ (Wikipedia). The relavent reaction is: $$\ce{4CrO3 ->[$\Delta \gt 197 \ \mathrm{^\circ C}$] 2Cr2O3 + 3O2}$$ Nonetheless, $\ce{Cr2O3}$ can also be easily reduced to elemental $\ce{Cr}$.
2. A large amount of chromic residues is discharged during the production of chromates by calcination of chromic ores. Statistics show that to produce a ton of sodium dichromate, the calcination process will discharge 2.5-3 tons of poisonous chromic residues. That chromic residues contain 3-7% of residual $\ce{Cr2O3}$, 8-11% of $\ce{Fe2O3}$, and 0.5-1.5% of water-soluble $\ce{Cr(VI)}$ (Ref.1). This reference has claimed a re-calcination of these poisonous chromic residues:
   Chronic residues are dried and crushed to 80-100 mesh, and mixed with chromic ore powder and sodium carbonate. Water is added to the the mixture to make it homogeneous (coke powder is also added as necessary), and then is heated in furnace at $\pu{1000-1200 ^\circ C}$. The main chemical reaction for this mixture is: $$\ce{4(FeO.Cr2O3) + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}.$$ Keep in mind that $\ce{(FeO.Cr2O3)}$ is representing $\ce{FeCr2O4}$ in this equation and $\ce{Na2CO3}$ replaces $\ce{NaOH}$ in OP's equation. Yet, the end results are the same.
3. The OP's second reaction, $\ce{Na2CrO4 + C ->[C, \Delta] CrO3}$ does not make any sense since oxidation of chromium does not change. I'd say it may be $\ce{Na2CrO4 + C ->[C, \Delta] Cr2O3}$. For example, the following conversion has been achied directly from $\ce{FeCr2O4}$: $$\ce{FeO.Cr2O3 + C -> Fe + Cr2O3 + CO}$$ Regardless, I also found following reaction to support my suggestion: $$\ce{2Na2CrO4 + 3C -> Cr2O3 + 2Na2O + 3CO}$$ Nevertheless, the reference 1 gives a different set of reactions for the conversion of $\ce{Na2CrO4}$: $$\ce{4Na2CrO4 + 6S + 7H2O -> 4Cr(OH)3 + 3Na2S2O3 + 2NaOH}$$ $$\ce{8Na2CrO4 + 6Na2S + 23H2O -> 8Cr(OH)3 + 3Na2S2O3 + 22NaOH}$$ $$\ce{2Cr(OH)3 ->[\pu{1200 ^\circ C}] Cr2O3 + 3H2O}.$$
4. For the 3rd reaction, $\ce{Al}$ is not the only reducing agent able to reduce $\ce{Cr2O3}$ to $\ce{Cr^\circ}$. The following reactions are listed in Ref.1: $$\ce{Cr2O3 + 3CO -> 2Cr + 3H2} \\ \ce{Cr2O3 + 3CO -> 2Cr + 3H2O} \\ \ce{Cr2O3 + 3C -> 2Cr + 3CO}$$ All of these reactions have been done at high temperatures.

Reference:

1. Qi-Jiang Situ, Ke-Ming Xu, Pei-Nian Huang, Xing-Qin Li, De-Han Zeng, Zhi-Fa Hu, Zhi-Quan Wen, "Re-calcination and extraction process for the detoxification and comprehensive utilization of chromium residues," United States Patent 1995, 5,395,601 (PDF).