Reduction Stage Mass Balance: Magnesium & MgCl2 Quantification

Hello! Welcome back.

In our last lesson, we explored the design and operational sequence of a Kroll batch reactor. We covered the how—the machinery, the inert atmosphere, and the critical control of temperature and pressure. Today, we shift from a qualitative description to a quantitative one.

This lesson focuses on answering the question: how much? You will learn to perform mass balance calculations for the reduction stage, quantifying the consumption of magnesium and the generation of the magnesium chloride byproduct. This is the foundation of process accounting. Just as a business must track every dollar, an industrial process must track every kilogram of material. These calculations are the first step in building a techno-economic model, directly linking the chemistry to the plant's operating costs and efficiency.

1. The Stoichiometric Recipe

The heart of any mass balance calculation is the balanced chemical equation. It provides the exact ratio of reactants consumed to products created. For the Kroll reduction, the equation is:

${\text{TiCl}}_4+2\text{Mg}\to \text{Ti}+2{\text{MgCl}}_2$

This equation tells us that 1 mole of titanium tetrachloride reacts with 2 moles of magnesium to produce 1 mole of titanium and 2 moles of magnesium chloride.

To convert these molar ratios into mass—the unit we care about for industrial production—we use the molar mass of each compound.

Let's walk through a simple, ideal calculation.

Problem: Assuming a 100% complete reaction, how much magnesium ($\text{Mg}$) is required to react with 1,000 kg of titanium tetrachloride (${\text{TiCl}}_4$), and how much titanium ($\text{Ti}$) and magnesium chloride (${\text{MgCl}}_2$) are produced?

Solution Steps:

1. Convert input mass to moles:
   ${\text{moles of TiCl}}_4=\frac{1,000,000\text{ g}}{189.68\text{ g/mol}}\approx 5272.0\text{ mol}$

2. Use stoichiometric ratios to find moles of other substances:

   • $\text{moles of Mg}=5272.0{\text{ mol TiCl}}_4\times \frac{2\text{ mol Mg}}{1{\text{ mol TiCl}}_4}=10544.0\text{ mol}$
   • $\text{moles of Ti}=5272.0{\text{ mol TiCl}}_4\times \frac{1\text{ mol Ti}}{1{\text{ mol TiCl}}_4}=5272.0\text{ mol}$
   • ${\text{moles of MgCl}}_2=5272.0{\text{ mol TiCl}}_4\times \frac{2{\text{ mol MgCl}}_2}{1{\text{ mol TiCl}}_4}=10544.0\text{ mol}$

3. Convert moles back to mass:

   • $\text{mass of Mg}=10544.0\text{ mol}\times 24.31\text{ g/mol}\approx 256.3\text{ kg}$
   • $\text{mass of Ti}=5272.0\text{ mol}\times 47.87\text{ g/mol}\approx 252.3\text{ kg}$
   • ${\text{mass of MgCl}}_2=10544.0\text{ mol}\times 95.21\text{ g/mol}\approx 1003.9\text{ kg}$

Notice that the total mass of reactants (1000 kg + 256.3 kg = 1256.3 kg) equals the total mass of products (252.3 kg + 1003.9 kg = 1256.2 kg), confirming the law of conservation of mass.

Now, try a quick practice problem yourself based on the image below.

Take a moment to calculate the answer. You should find that for 2.50 g of TiCl₄, approximately 2.51 g of MgCl₂ are produced.

2. Reality Check: Excess Reactants and Incomplete Conversion

Industrial processes are not ideal. To maximize the conversion of the expensive, purified ${\text{TiCl}}_4$, operators use more magnesium than the stoichiometric "recipe" calls for. This is known as using an excess reactant.

As the text states, the surplus of magnesium is used "to assure complete reduction of the titanium tetrachloride". By flooding the reactor with the reducing agent, the process designers ensure that as much ${\text{TiCl}}_4$ as possible finds a magnesium atom to react with. This pushes the reaction towards completion and maximizes the yield of titanium from the valuable feedstock.

However, even with excess magnesium, the conversion is never 100%. Some ${\text{TiCl}}_4$ may be lost through venting, or some may fail to react. This is captured by the conversion rate.

The Mintek experiments show a conversion rate of around 80-82% is more realistic. The reasons are practical and highlight the challenges of industrial chemistry:

• Loss of gaseous ${\text{TiCl}}_4$ during pressure venting.
• Entrapment of unreacted magnesium in the solid sponge, making it unavailable.
• Losses of fine titanium particles during handling.

Let's incorporate these factors into a more realistic problem.

Problem: We want to produce 10,000 kg of titanium sponge in a batch. A typical industrial process uses 25% excess magnesium and achieves an 85% conversion rate of ${\text{TiCl}}_4$ to $\text{Ti}$. Calculate the required mass of ${\text{TiCl}}_4$ and $\text{Mg}$, and the mass of ${\text{MgCl}}_2$ produced.

Solution Steps:

1. Calculate the theoretical ${\text{TiCl}}_4$ required for the target Ti output (Ideal Case):

   • First, find moles of Ti desired:
     $\frac{10,000,000\text{ g Ti}}{47.87\text{ g/mol}}\approx 208,900\text{ mol Ti}$
   • From stoichiometry, moles ${\text{TiCl}}_4$ = moles $\text{Ti}$.
   • Theoretical mass of ${\text{TiCl}}_4$:
     $208,900\text{ mol}\times 189.68\text{ g/mol}\approx 39,623{\text{ kg TiCl}}_4$

2. Adjust for Real-World Conversion Rate:

   • The 39,623 kg is the amount that must successfully react. Since only 85% of the input reacts, we need to feed more.
   • Required ${\text{TiCl}}_4$ input:
     $\frac{39,623\text{ kg}}{0.85}\approx {\text{46,615 kg TiCl}}_4$

3. Calculate Required Magnesium with Excess:

   • First, find the stoichiometric Mg required for the actual amount of ${\text{TiCl}}_4$ fed:
     ${\text{moles of TiCl}}_4\text{ fed}=\frac{46,615,000\text{ g}}{189.68\text{ g/mol}}\approx 245,760\text{ mol}$
     $\text{moles of Mg}=245,760{\text{ mol TiCl}}_4\times 2=491,520\text{ mol Mg}$
     $\text{stoichiometric mass of Mg}=491,520\text{ mol}\times 24.31\text{ g/mol}\approx 11,949\text{ kg Mg}$
   • Add the 25% excess:
     $\text{Required Mg input}=11,949\text{ kg}\times 1.25\approx \text{14,936 kg Mg}$

4. Calculate ${\text{MgCl}}_2$ Byproduct:

   • The amount of byproduct is based on the amount of ${\text{TiCl}}_4$ that actually reacted.
   • Moles of ${\text{TiCl}}_4$ reacted: $245,760\text{ mol}\times 0.85=208,900\text{ mol}$
   • Moles of ${\text{MgCl}}_2$ produced: $208,900\text{ mol}\times 2=417,800\text{ mol}$
   • Mass of ${\text{MgCl}}_2$ produced:
     $417,800\text{ mol}\times 95.21\text{ g/mol}\approx {\text{39,780 kg MgCl}}_2$

This more complex calculation gives you a much better estimate of the material requirements for an actual plant, forming the basis for your cost model.

3. A Look at an Industrial Batch

Let's see how our calculations stack up against published data from a real industrial operation.

The following reading describes the OREMET process, a large-scale modification of the Kroll process.

The OREMET process data provides a fantastic opportunity to check our understanding. Let's analyze their numbers (using 1 lb = 0.453592 kg):

• Inputs:
  • TiCl₄: 46,000 - 47,000 lbs (approx. 20,865 - 21,319 kg)
  • Mg: 16,000 - 17,000 lbs (approx. 7,257 - 7,711 kg), stated as "25 percent in excess of stoichiometric requirements."
• Outputs:
  • Ti Sponge: 13,000 - 14,000 lbs (approx. 5,897 - 6,350 kg)
  • MgCl₂ Tapped: 46,000 lbs (approx. 20,865 kg)

Verification Exercise:
Let's see if their "25% excess" claim holds up.
Using the average input of 46,500 lbs (21,092 kg) of ${\text{TiCl}}_4$:

1. Moles ${\text{TiCl}}_4$ = $\frac{21,092,000\text{ g}}{189.68\text{ g/mol}}\approx 111,200\text{ mol}$
2. Stoichiometric Moles Mg = $111,200\times 2=222,400\text{ mol}$
3. Stoichiometric Mass Mg = $222,400\text{ mol}\times 24.31\text{ g/mol}\approx 5,406\text{ kg}$ (or ~11,918 lbs)
4. Adding 25% excess: $5,406\text{ kg}\times 1.25\approx 6,758\text{ kg}$ (or ~14,900 lbs)

The OREMET input is 16,000-17,000 lbs of Mg. This is slightly more than our calculated requirement of ~14,900 lbs, suggesting their "25% excess" figure is a rounded number and they may use closer to 35-40% excess in practice. This is a crucial insight—published numbers often simplify the complex reality. As an investor or operator, you would need to dig deeper to understand the true consumption factor.

Similarly, we can calculate the theoretical titanium yield from 46,500 lbs of ${\text{TiCl}}_4$:

• Theoretical Ti Mass = $111,200\text{ mol}\times 47.87\text{ g/mol}\approx 5,323\text{ kg}$ (or 11,735 lbs).

The actual output is 13,000-14,000 lbs of sponge. The sponge, however, is not pure titanium; it contains residual Mg and MgCl₂ before purification. This accounts for the higher mass.

Conclusion

Today, we have translated the chemistry of the Kroll reduction into the language of industrial operations. You are now equipped to calculate the material inputs and outputs for this critical process step, accounting for the practical realities of excess reactants and incomplete conversion.

Key Takeaways:

• Mass balance calculations are founded on the stoichiometric ratios of the balanced chemical equation: ${\text{TiCl}}_4+2\text{Mg}\to \text{Ti}+2{\text{MgCl}}_2$.
• Industrial practice deviates from theory: excess magnesium (e.g., 25% or more) is used to maximize the conversion of the primary feedstock, ${\text{TiCl}}_4$.
• Real-world conversion rates are below 100% (typically 80-90%) due to physical process limitations, and this must be factored into any serious calculation.
• By quantifying material flows, you are building the essential data needed for a techno-economic analysis, estimating raw material costs, and evaluating process efficiency.

The output of the reactor, as we've seen, is not a finished product. It's a crude "sponge" contaminated with significant amounts of unreacted magnesium and magnesium chloride. In our next lesson, we will examine the crucial purification steps—vacuum distillation and leaching—used to remove these impurities and produce the high-purity titanium sponge required by the market.