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Chapter 11: Problem 38

Titanium is produced from its principal ore, rutile, \(\mathrm{TiO}_{2}(s)\), by the two-step process described by (1) \(\mathrm{TiO}_{2}(s)+2 \mathrm{Cl}_{2}(g)+2 \mathrm{C}(s) \rightarrow\) $$ \mathrm{TiCl}_{4}(g)+2 \mathrm{CO}(g) $$ (2) \(\mathrm{TiCl}_{4}(g)+2 \mathrm{Mg}(s) \rightarrow \mathrm{Ti}(s)+2 \mathrm{MgCl}_{2}(s)\) How many kilograms of titanium can be produced from \(4.10 \times 10^{3}\) kilograms of \(\mathrm{TiO}_{2}(s) ?\)

Short Answer

Expert verified
2,458 kg of titanium can be produced.

Step by step solution

01

Convert Kilograms to Moles

First, we need to convert the mass of \( \mathrm{TiO}_{2}(s) \) to moles. The molar mass of \( \mathrm{TiO}_{2} \) is calculated as follows: \((1 \times 47.87) + (2 \times 16.00) = 79.87\, \text{g/mol} \). Given the mass is \( 4.10 \times 10^{3} \) kg or \( 4.10 \times 10^{6} \) g, the number of moles of \( \mathrm{TiO}_{2} \) is:\[ n = \frac{4.10 \times 10^{6}\, \text{g}}{79.87\, \text{g/mol}} = 5.135 \times 10^{4} \text{ moles} \].
02

Determine Moles of Titanium Produced

The stoichiometry of the first reaction indicates that 1 mole of \( \mathrm{TiO}_{2} \) produces 1 mole of \( \mathrm{TiCl}_{4} \), which is then used in a 1:1 ratio in the second reaction to produce 1 mole of \( \mathrm{Ti} \). Therefore, the number of moles of \( \mathrm{Ti} \) produced is equal to the number of moles of \( \mathrm{TiO}_{2} \): \[ n_{\mathrm{Ti}} = 5.135 \times 10^{4} \text{ moles} \].
03

Convert Moles of Titanium to Kilograms

Now, convert the moles of titanium to mass. The molar mass of \( \mathrm{Ti} \) is \( 47.87 \text{ g/mol} \). The mass of titanium produced is calculated as :\[ m_{\mathrm{Ti}} = 5.135 \times 10^{4}\, \text{moles} \times 47.87\, \text{g/mol} = 2.458 \times 10^{6}\, \text{g} = 2.458 \times 10^{3}\, \text{kg} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Understanding molar mass is crucial for solving problems in chemical stoichiometry. The molar mass of a substance refers to the mass of one mole of its molecules or formula units. It is expressed in units of grams per mole (g/mol). To find the molar mass, we sum up the atomic masses of all the elements in a molecule.
For example, let's consider rutile, \(\text{TiO}_2\). The molar mass is calculated by adding the atomic mass of titanium (Ti), which is 47.87 g/mol, to twice the atomic mass of oxygen (O), which is 16.00 g/mol each. Therefore, the molar mass of \(\text{TiO}_2\) is:
  • \( 47.87 + 2 \times 16.00 = 79.87 \, \text{g/mol} \)
These calculations are essential because they allow us to convert between mass and moles, facilitating further stoichiometric calculations.
Stoichiometric Calculations
Stoichiometric calculations aim to determine the quantitative relationships between reactants and products in a chemical reaction. This involves using the balanced chemical equation to find the proportion, or ratio, in which the substances react. The ratios are derived from the coefficients in the chemical equation.
For the task at hand, two reactions are involved:
  • The conversion of \(\text{TiO}_2\) to \(\text{TiCl}_4\) and \(\text{TiCl}_4\) to Ti.
Each of these reactions shows a one-to-one correspondence between reactants and products. This means that 1 mole of \(\text{TiO}_2\) yields 1 mole of \(\text{TiCl}_4\), which then yields 1 mole of Ti. Such relationships allow us to calculate how much titanium can be produced from a given amount of \(\text{TiO}_2\).
To perform this calculation, first convert the mass of \(\text{TiO}_2\) to moles using its molar mass, then apply the stoichiometric coefficients from the balanced equations.
Chemical Reactions
Chemical reactions describe how substances transform into new products. The balanced chemical equations provide a symbolic representation of these transformations. Each reactant and product in a reaction has a specific formula and a corresponding coefficient to fulfill the law of conservation of mass.
In our example, the processes involve:
  • The first reaction transforms \(\text{TiO}_2\) into \(\text{TiCl}_4\) using chlorine and carbon.
  • The second reaction converts \(\text{TiCl}_4\) into titanium using magnesium.
The coefficients, like the ones from the balanced equations of these two reactions, are crucial. They guide us in performing accurate stoichiometric calculations by showing us how moles of reactants convert into moles of products.
Understanding the nature of these transformations helps predict the feasible outcomes of the reactions, facilitating precise measurement and material usage in practical applications.

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Most popular questions from this chapter

An ore is analyzed for its lead content as follows. A sample is dissolved in water; then sodium sulfate is added to precipitate the lead as lead(II) sulfate, \(\mathrm{PbSO}_{4}(s) .\) The net ionic equation for the reaction is $$ \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{PbSO}_{4}(s) $$ It was found that \(13.73\) grams of lead(II) sulfate were precipitated from a sample of ore having a mass of \(53.92\) grams. How many grams of lead are there in the sample? What is the mass percentage of lead in the ore?

II-53. Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)\), an ore used in the production of aluminum, can be synthesized by the reaction described by the equation \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{NaOH}(l)+\mathrm{HF}(g) \stackrel{\text { High } T}{\longrightarrow}\) \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad(\) unbalanced \()\) (a) Balance this equation. (b) If \(10.0\) kilograms of \(\mathrm{Al}_{2} \mathrm{O}_{3}(s), 50.00\) kilograms of \(\mathrm{NaOH}(l)\), and \(50.0\) kilograms of \(\mathrm{HF}(g)\) react completely, how many kilograms of cryolite will be produced? (c) Which reactants will be in excess and how many kilograms of each of these reactants will remain?

A hydrated form of copper(II) sulfate, \(\mathrm{CuSO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O}(s)\), is heated to drive off all the waters of hydration. If we start with \(9.40\) grams of hydrated salt and have \(5.25\) grams of anhydrous \(\mathrm{CuSO}_{4}(s)\) after heating, find the number of water molecules, \(n\), associated with each CuSO \(_{4}\) formula unit.

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, \(\mathrm{KClO}_{3}(s)\). The equation for the reaction is $$ 2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ Calculate how many grams of \(\mathrm{O}_{2}(g)\) can be produced from heating \(10.0\) grams of \(\mathrm{KClO}_{3}(s)\).

A website promoting the use of alternative energy vehicles and hybrid technologies claims that, "A typical automobile in the USA uses about 500 gallons of gasoline per year, producing about 5 tons of carbon dioxide." Does this statement make sense? Make your own estimate assuming that the primary ingredient in gasoline is octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), which has a density of \(0.7 \mathrm{~g} \cdot \mathrm{m} \mathrm{L}^{-1}\)
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