Q22.65CP - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: 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:

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

Step 2: 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%.

Step 3: 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:

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

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