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Aluminum is widely distributed throughout the world in the form of aluminosilicates. What property of these minerals prevents them from being a source of aluminum?

Short Answer

Expert verified
The strong and stable bonds in aluminosilicates make aluminum extraction energy-intensive and costly.

Step by step solution

01

Understand the Composition of Aluminosilicates

Aluminosilicates are minerals composed of aluminum, silicon, and oxygen. These minerals form very stable structures due to strong bonds between the elements.
02

Identify the Bond Characteristics

The aluminum in aluminosilicates is bound in a crystal lattice with silicon and oxygen. This bonding is quite strong and energy-intensive to break, making it difficult to extract aluminum.
03

Consider the Extraction Process

The extraction of aluminum from aluminosilicates would require breaking these strong bonds. This process is not only energy-intensive but also costlier compared to extracting aluminum from more easily processed minerals like bauxite.
04

Compare with Other Aluminum Sources

Compare the difficulty of extracting aluminum from aluminosilicates versus from bauxite. Bauxite is preferred for aluminum extraction because its chemical composition allows for a more straightforward and less energy-intensive extraction process.

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Key Concepts

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

aluminosilicates
Aluminosilicates are a group of minerals composed of aluminum, silicon, and oxygen. They are found widely distributed across the Earth's crust. These minerals are known for their incredibly stable structures, resulting from strong bonds between aluminum, silicon, and oxygen atoms. Because of this stability, extracting aluminum from aluminosilicates requires a significant amount of energy. The chemical bonds within aluminosilicates are so strong that breaking them is not practical for aluminum extraction purposes. Understanding the robust nature of these bonds explains why aluminosilicates are not a common source of aluminum.
crystal lattice
A crystal lattice is a repeating three-dimensional structure formed by atoms, ions, or molecules. In the case of aluminosilicates, the aluminum atoms are embedded within a lattice of silicon and oxygen atoms. This lattice framework provides a highly ordered structure, contributing to the mineral’s stability. Due to the tightly bound atoms in a crystal lattice, a substantial amount of energy is required to disrupt this arrangement to extract individual elements like aluminum. The strength and stability of the crystal lattice are key factors making aluminosilicates impractical for aluminum extraction compared to more easily broken down minerals.
bauxite
Bauxite is the principal ore of aluminum and is preferred for aluminum extraction. It primarily contains aluminum hydroxide minerals, which are much easier to process compared to aluminosilicates. The chemical bonds in bauxite are weaker, allowing for a more straightforward and less energy-intensive extraction process. The extraction involves the Bayer process, where bauxite is dissolved in sodium hydroxide, separating aluminum oxide from impurities. The ease and efficiency of extracting aluminum from bauxite make it the most economically viable source, which is why it is extensively used over more complex minerals like aluminosilicates.

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Most popular questions from this chapter

22.79 Ores with as little as \(0.25 \%\) by mass of copper are used as sources of the metal. (a) How many kilograms of such an ore would be needed to construct another Statue of Liberty, which contains \(2.0 \times 10^{3}\) Ib of copper? (b) If the mineral in the ore is chalcopyrite (CuFeS, ), what is the mass \(\%\) of chalcopyrite in the ore?

World production of chromite \(\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right),\) the main ore of chromium, was \(1.97 \times 10^{7}\) metric tons in 2006 . To isolate chromium, a mixture of chromite and sodium carbonate is heated in air to form sodium chromate, iron(III) oxide, and carbon dioxide. The sodium chromate is dissolved in water, and this solution is acidified with sulfuric acid to produce the less soluble sodium dichromate. The sodium dichromate is filtered out and reduced with carbon to produce chromium(III) oxide, sodium carbonate, and carbon monoxide. The chromium(III) oxide is then reduced to chromium with aluminum metal. (a) Write balanced equations for each step. (b) What mass of chromium (in \(\mathrm{kg}\) ) could be prepared from the 2006 world production of chromite?

Several transition metals are prepared by reduction of the metal halide with magnesium. Titanium is prepared by the Kroll method, in which the ore (ilmenite) is converted to the gaseous chloride, which is then reduced to Ti metal by molten \(\mathrm{Mg}\) (see the discussion on the isolation of magnesium in Section 22.4 ). Assuming yiclds 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?

The following steps are unbalanced half-reactions involved in the nitrogen cycle. Balance each half-reaction to show the number of electrons lost or gained, and state whether it is an oxidation or a reduction (all occur in acidic conditions): (a) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NO}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g)\) (c) \(\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (d) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (c) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NO}_{3}^{-}(a q)\)

Farmers use ammonium sulfate as a fertilizer. In the soil. nitrifying bacteria oxidize \(\mathrm{NH}_{4}^{+}\) to \(\mathrm{NO}_{3}^{-},\) a groundwater contaminant that causes methemoglobinemia ("blue baby" syndrome). The World Health Organization standard for maximum \(\left[\mathrm{NO}_{3}^{-}\right]\) in groundwater is \(45 \mathrm{mg} / \mathrm{L}\). A farmer adds \(210 . \mathrm{kg}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) to a field and \(37 \%\) is oxidized to \(\mathrm{NO}_{3}^{-}\). What is the groundwater \(\left[\mathrm{NO}_{3}^{-}\right]\) (in \(\mathrm{mg} / \mathrm{L}\) ) if \(1000 . \mathrm{m}^{3}\) of the water is contaminated?

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