Question

An important use for magnesium is to make titanium. In the Kroll process, magnesium reduces titanium(IV) chloride to elemental titanium in a sealed vessel at 800°C. Write a balanced chemical equation for this reaction.

What mass of magnesium is needed, in theory, to produce 100 kg of titanium from titanium(IV) chloride?

Short answer

The balanced chemical equation is $2\mathrm{Mg}+{\mathrm{TiCl}}_4\to \mathrm{Ti}+2{\mathrm{MgCl}}_2$. The mass of magnesium is 102 kg.

Step-by-step solution

Balanced chemical equation

It is given that magnesium reduces titanium (IV) chloride to elemental titanium. So, the balanced chemical equation is $2\mathrm{Mg}+{\mathrm{TiCl}}_4\to \mathrm{Ti}+2{\mathrm{MgCl}}_2$.

Step 2: Calculation of the number of moles of titanium

Given the mass of titanium, ${\mathrm{m}}_{\mathrm{Ti}}=800\mathrm{kg}=800\times {10}^3\mathrm{g}$

We need to find the number of moles of titanium to calculate the number of moles of magnesium, followed by the mass of magnesium.

The molar mass of titanium, ${\mathrm{M}}_{\mathrm{Ti}}=47.867{\mathrm{gmol}}^{-1}$

The number of moles of titanium,

$\begin{array}{c}{\mathrm{n}}_{\mathrm{Ti}}=\frac{{\mathrm{m}}_{\mathrm{Ti}}}{{\mathrm{M}}_{\mathrm{Ti}}}\\ =\frac{100\times {10}^3\mathrm{g}}{47.867{\mathrm{gmol}}^{-1}}\\ =2.089\times {10}^3\mathrm{mol}\end{array}$

The number of moles of titanium is $2.089\times {10}^3\mathrm{mol}$.

Calculation of number of moles of magnesium

According to the chemical equation, 2 moles of magnesium produces 1 mole of titanium. So, the number of moles of magnesium will be twice that of titanium.

The number of moles of magnesium, ${\mathrm{n}}_{\mathrm{Mg}}=4.178\times {10}^3\mathrm{mol}$.

Calculation of the mass of magnesium

The mass of magnesium is calculated by the equation as follows:

${\mathrm{n}}_{\mathrm{Mg}}=\frac{{\mathrm{m}}_{\mathrm{Mg}}}{{\mathrm{M}}_{\mathrm{Mg}}}$, where ${\mathrm{m}}_{\mathrm{Mg}}$ is the mass of magnesium, and ${\mathrm{M}}_{\mathrm{Mg}}$is the molar mass of magnesium.

$\begin{array}{c}4.178\times {10}^3\mathrm{mol}=\frac{{\mathrm{m}}_{\mathrm{Mg}}}{24.305{\mathrm{gmol}}^{-1}}\\ {\mathrm{m}}_{\mathrm{Mg}}=4.178\times {10}^3\mathrm{mol}\times 24.305{\mathrm{gmol}}^{-1}\\ =1.015\times {10}^5\mathrm{g}\\ =102\mathrm{kg}\end{array}$

Hence, the mass of magnesium is 102 kg.