91Ó°ÊÓ

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:

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

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:

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

Therefore, the total yield is 78%.

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:

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

Let’s find the mass of metal:

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

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