Question

Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)

Short answer

The maximum mass of \(TiCl4\) that can be obtained is 30 g.

Step-by-step solution

Convert given mass to moles

Molar mass of \(TiO2\) is about 79.9 g/mol. Moles of \(TiO2\) = 35 g / 79.9 g/mol = 0.437 mol. Molar mass of \(Cl2\) is about 71 g/mol. Moles of \(Cl2\) = 45 g / 71 g/mol = 0.633 mol. Molar mass of \(C\) is about 12 g/mol. Moles of \(C\) = 11 g / 12 g/mol = 0.917 mol.

Identify the limiting reactant

Looking at the balanced chemical equation, for every 3 moles of \(TiO2\), 6 moles of \(Cl2\) and 4 moles of \(C\) are required. So, normalized for \(TiO2\): \(TiO2\) = 0.437 mol, \(Cl2\) = 0.633 mol / 2 = 0.316 mol, \(C\) = 0.917 mol / 1.33 = 0.69 mol. In conclusion, \(Cl2\) is the limiting reactant because it has the smallest mole quantity.

Calculate the maximum mass of \(TiCl4\)

The balanced equation tells us that for every 6 moles of \(Cl2\), 3 moles of \(TiCl4\) can be formed. Therefore, moles of \(TiCl4\) = moles of \(Cl2\) / 2 = 0.316 mol / 2 = 0.158 mol. The molar mass of \(TiCl4\) is 189.7 g/mol. So, the maximum mass of \(TiCl4\) is = moles of \(TiCl4\) * molar mass of \(TiCl4\) = 0.158 mol * 189.7 g/mol = 30 g.

Key concepts

Limiting Reactants

In chemical reactions, the concept of limiting reactants is crucial for determining the amount of product that can be formed. In any reaction, the limiting reactant is the substance that runs out first, halting the reaction because there’s no more of it to continue reacting. This is an essential concept because it dictates the maximum quantity of the product that can be created. For the reaction \[3 \mathrm{TiO}_{2} + 4 \mathrm{C} + 6 \mathrm{Cl}_{2} \rightarrow 3 \mathrm{TiCl}_{4} + 2 \mathrm{CO}_{2} + 2 \mathrm{CO}\] we need to consider the amounts of \(\mathrm{TiO}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{C}\) provided. By converting these into moles, we can compare their ratios to the balanced equation and identify which reactant limits the reaction. By normalizing the mole values, \(\mathrm{Cl}_{2}\) was found to be the limiting reactant as it offered the smallest reactant-to-product ratio. Thus, \(\mathrm{Cl}_{2}\) restricts the reaction to form the maximum possible mass of \(\mathrm{TiCl}_{4}\). This highlights why identifying the limiting reactant is essential for accurate stoichiometric calculations.

Molar Mass Calculation

Molar mass calculations play a vital role in stoichiometry, helping convert mass of substances to moles, which are necessary for balanced chemical equations. Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). To determine the amount of each reactant, we need to calculate the molar masses:

• \(\mathrm{TiO}_{2}\): Approximately 79.9 g/mol
• \(\mathrm{Cl}_{2}\): Approximately 71 g/mol
• \(\mathrm{C}\): 12 g/mol

By using these values, we can convert the mass of each reactant to moles. For example, to convert \(35 \mathrm{g \ of \ TiO}_{2}\) to moles, divide by its molar mass: \(35 \, \text{g} / 79.9 \, \text{g/mol} = 0.437 \, \text{mol}\). This conversion is crucial in understanding the extent of chemical reactions and finding out which reactant is the limiting reactant. Without accurate molar mass calculations, our stoichiometric analysis would be flawed.

Chemical Reactions

Chemical reactions describe the processes through which substances interact to form new products. Understanding them requires the ability to decipher balanced chemical equations. The equation provided in this exercise \[3 \mathrm{TiO}_{2} + 4 \mathrm{C} + 6 \mathrm{Cl}_{2} \rightarrow 3 \mathrm{TiCl}_{4} + 2 \mathrm{CO}_{2} + 2 \mathrm{CO}\] denotes a synthesis reaction where

• 3 moles of \(\mathrm{TiO}_{2}\)
• 4 moles of \(\mathrm{C}\)
• 6 moles of \(\mathrm{Cl}_{2}\)

combine to produce 3 moles of \(\mathrm{TiCl}_{4}\), alongside carbon dioxide and carbon monoxide gases. To solve stoichiometric problems like this, we need to interpret the equation correctly, recognizing amounts of reactants and products. Each coefficient in the equation indicates the number of moles of each substance involved, serving as a critical guide for calculating reactant consumption and product formation. This understanding of chemical reactions allows us to predict the outcomes of the reaction efficiently.