Question

The chemical formula of bauxite is:
A.\[A{l_2}{O_3}\]
B.\[A{l_2}{O_3}.2{H_2}O\]
C.\[N{a_3}Al{F_6}\]
D.\[A{l_2}{0_6}\]

Answer 1

Hint: Chemical formula is a way to write a substance using the chemical symbol and number subscripts to mention the number of the atoms into that chemical compound. Bauxite is an inorganic compound. It is an ore of aluminum.

Complete step by step answer:
The main source of ore and mineral is the earth. Metals and their components are found in minerals. The ore is the mineral from which metals can be extracted conventionally and cheaply.
In the case of the ore, the percentage of metal is greater than other minerals. For example, the ore of aluminum is bauxite. From bauxite, aluminum can be extracted cheaply and profitably. But aluminum is also found in clay, but it is not possible to extract aluminum from clay cheaply and profitably. That is why bauxite is the ore of aluminum.
The chemical formula of bauxite is, \[A{l_2}{O_3}.2{H_2}O\] .

So, the correct answer is B.

Additional information:
The hall-Heroult process is a major industrial process that is used for the smelting of \[{\text{Al}}{\text{.}}\]In the Hall-Heroult process, alumina is dissolved in molten cryolite \[{\text{N}}{{\text{a}}_{\text{3}}}{\text{[Al}}{{\text{F}}_{\text{6}}}{\text{]}}\] by using Bauxite that is the chief ore of \[{\text{Al}}{\text{.}}\]This molten salt bath is then electrolyzed in a purpose-built cell.
In the Hall-Heroult process, pure aluminum oxide or alumina \[{\text{(A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{)}}\] is combined with \[{\text{Na}}{{\text{F}}_{\text{2}}}\] or \[{\text{N}}{{\text{a}}_{\text{3}}}{\text{[Al}}{{\text{F}}_{\text{6}}}{\text{]}}\] which results in the lowering of the melting point of the mixture, and the ability to conduct electricity increases. A steel vessel that is lined with carbon and graphite rods is used in this process.
The carbon lining is used as the cathode whereas the graphite rods are used as anodes. Upon passage of electricity through the electrolytic cell, oxygen is produced at the anode. This oxygen reacts with the carbon of the anode, to form carbon monoxide and carbon dioxide.
Aluminum ions are produced at the cathode from the \[{\text{(A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{)}}\] and sink to the bottom as these ions are heavier than the cryolite solution. The oxygen of \[{\text{(A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{)}}\] reacts with carbon dioxide with the graphite rods.
The reaction at the cathode is:
\[{\text{A}}{{\text{l}}^{{\text{3 + }}}}{\text{ + 3}}{{\text{e}}^{\text{ - }}} \to {\text{Al(l)}}\]
The reaction at the anode:
\[{\text{C(s) + }}{{\text{O}}^{{\text{2 - }}}} \to {\text{CO(g) + 2}}{{\text{e}}^{\text{ - }}}\]
\[{\text{C(s) + 2}}{{\text{O}}^{{\text{2 - }}}} \to {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g) + 4}}{{\text{e}}^{\text{ - }}}\]
The overall reaction is:
\[{\text{2A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{ + 3C}} \to {\text{4Al + 3C}}{{\text{O}}_{\text{2}}}\]

Note:Smelting is the process of obtaining a metal by heating its ore above the melting point in the presence of oxidizing agents such as air or reducing agents such as coke. Most ores are oxides and this process removes the oxygen from the oxide, leaving behind the metal. Even then, if the metal is still not pure, so then another process is needed.