Question

Several transition metals are prepared by reduction of the metal halide with magnesium. Titanium is prepared by the Kroll method, in which ore (ilmenite) is converted to the gaseous chloride, which is then reduced to metal by molten Mg(see p. 1008). Assuming yields of 84% for step 1 and 93% for step 2, and an excess of the other reactants, what mass of Ti metal can be prepared from 21.5 metric tons of ilmenite?

Short answer

$5.3$ton mass of $\mathrm{Ti}$ can be prepared from $21.5$ metric tons of ilmenite.

Step-by-step solution

 Concept Introduction

The idealised formula for ilmenite is ${\mathbf{\text{FeTiO}}}_{\mathbf{\text{3}}}$which is a titanium-iron ore. The method for isolating titanium can be described by using equations below:

$$\begin{gathered} {\mathbf{\text{2FeTiO}}}_{\mathbf{\text{3}}}{\mathbf{\text{(s) + 7Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g) + 6C(s)}}\mathbf{\to }{\mathbf{\text{2TiCl}}}_{\mathbf{\text{4}}}{\mathbf{\text{(l) + 2FeCl}}}_{\mathbf{\text{3}}}\mathbf{\text{(s) + 6CO(g)}} \\ {\mathbf{\text{TiCll}}}_{\mathbf{\text{4}}}\mathbf{\text{(l) + 2Mg(l)}}\mathbf{\to }{\mathbf{\text{Ti(s) + 2MgCl}}}_{\mathbf{\text{2}}}\mathbf{\text{(l)n}} \end{gathered}$$

 Calculate the mass of Ti metal

We can determine the mass of Ti metal which can be extracted from $21.5$metric tonnes of ilmenite by assuming yields of 84% for process one and 93% for process two. The total process yield can be determined by multiplying the outputs of each process step:

$$\begin{gathered} yield{}_{total}=yield1\times yield{}_2 \\ yield{}_{total}=0.84\times 0.93 \\ yield{d}_{total}=0.78=78\% \end{gathered}$$

Therefore, the total yield is 78%.

 Calculate the chemical amount of ilmenite

We can figure out the biochemical quantity of ilmenite, then the chemical proportion of metal (assuming a yield of less than ), and lastly the mass of metal:

$$\begin{gathered} m\left(FeTi{O}_3\right)=21.5t=21.5\times {10}^{6}g \\ M\left(FeTi{O}_3\right)=151.71gmo{l}^{-1} \\ n\left(FeTi{O}_3\right)=\frac{m\left(FeTi{O}_3\right)}{M\left(FeTi{O}_3\right)} \\ n\left(FeTi{O}_3\right)={\frac{21.5\times {10}^{6}g}{151.71gmol}}^{-1} \\ n\left(FeTi{O}_3\right)=141718mol \end{gathered}$$

Let’s find the mass of metal:

$$\begin{gathered} n\left(Ti\right)=n\left(FeTi{O}_3\right)m\left(Ti\right) \\ =n\left(Ti\right)\times M\left(Ti\right)\times yield{d}_{total} \\ m\left(Ti\right)=141718mol\times 47.867gmo{l}^{-1}\times 0.78 \\ m\left(Ti\right)=5290879g \\ =5.3tons \end{gathered}$$

Therefore, the mass of Ti metal is $5.3$tons