I am looking for a process in which reduces ilmenite to pure titanium with the highest yield possible. We are to use a process of our choice or one that we have created in conjunction with FactSage to calculate the yield of titanium.

the materials we have available are:

  • Pure Metals
    • $\ce{Al, Ca, Mg, Na, Si}$
  • Pure Non-metals
    • $\ce{Ar, C, Cl2, H2, N2, O2, S }$
  • Molecules
    • $\ce{CH4, CO, CO2, HCl, H2O, H2S, NH4}$

There are a few process that come to mind such as reduction using Magnesium, $\ce{H2SO4}$, high temperature $\ce{CO2}$, but any help on which would be the most efficient combined with the highest yield would be great.

  • 1
    $\begingroup$ @Micheal it would be better, if you would use the mhchem package to format chemistry, rather than HTML. $\endgroup$ Commented Nov 17, 2014 at 7:38
  • $\begingroup$ Thermite Reduction is easiest. $\endgroup$
    – Lighthart
    Commented Nov 17, 2014 at 20:01

1 Answer 1


Ilmenite mainly consists of $\ce{TiO2}$ and iron oxides. The idealized formula of ilmenite is approximately $\ce{FeTiO3}$.

Note that the aluminothermic process would yield ferrotitanium alloys, and the reduction with carbon would yield titanium carbide.

The titanium ore can be converted to titanium tetrachloride according to $$\ce{TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO}$$ or $$\ce{2FeTiO3 + 7Cl2 + 6C -> 2TiCl4 + 2FeCl3 + 6CO}$$ (If a large content of iron oxides causes problem, it can be partially removed first by partial reduction with carbon.)

Titanium tetrachloride (boiling point: 136.5 °C) can be purified by fractional distillation.

Finally, titanium tetrachloride can be reduced to titanium sponge with magnesium (Kroll process): $$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$

The remaining $\ce{Mg}$ and $\ce{MgCl2}$ can be removed with hydrochloric acid.


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