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Question: Anhydrous \({\text{AlC}}{{\text{l}}_3}\) is prepared by: A. Reaction of \({\text{HCl}}\) and \({\t...

Anhydrous AlCl3{\text{AlC}}{{\text{l}}_3} is prepared by:
A. Reaction of HCl{\text{HCl}} and Al{\text{Al}} metal.
B. Reaction of dry HCl{\text{HCl}} and heated Al{\text{Al}} metal.
C. Passing HCl{\text{HCl}} gas over heated Al{\text{Al}} metal
D. Reaction of Al2O3{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}} with dil. HCl{\text{HCl}}.

Explanation

Solution

Aluminium chloride is produced by an exothermic reaction of aluminium and chlorine. The produced aluminium chloride is anhydrous in nature.

Complete step by step answer:
Anhydrous aluminium chloride is prepared by the reaction of dry hydrochloric acid and heated aluminium metal. The reaction is exothermic.
The reaction is,
6Al+6HCl2AlCl3+H2{\text{6Al}} + {\text{6HCl}} \to {\text{2AlC}}{{\text{l}}_{\text{3}}} + {{\text{H}}_{\text{2}}}
Thus, anhydrous AlCl3{\text{AlC}}{{\text{l}}_3} is prepared by reaction of dry HCl{\text{HCl}} and heated Al{\text{Al}} metal.
Thus, the correct option is (B) reaction of dry HCl{\text{HCl}} and heated Al{\text{Al}} metal.

Additional Information:
Properties of aluminium chloride are:
1. Aluminium chloride has low melting point and low boiling point.
2. Aluminium chloride is a poor conductor of electricity in its molten state.
3. The sublimation temperature of aluminium chloride is 180C{180^ \circ }{\text{C}}.
4. Aluminium chloride is a good Lewis acid.
5. Aluminium chloride is anhydrous in nature.
6. Aluminium chloride is non-explosive, non-flammable and corrosive.
7. On contact with water, aluminium chloride reacts violently.

Note:
The other ways in which aluminium chloride can be prepared are as follows:
Reaction of aluminium metal with chlorine gas. The reaction is,
2Al+3Cl32AlCl3{\text{2Al}} + {\text{3C}}{{\text{l}}_{\text{3}}} \to {\text{2AlC}}{{\text{l}}_{\text{3}}}
Displacement reaction in which aluminium metal displaces copper in copper chloride.
2Al+3CuCl22AlCl3+3Cu{\text{2Al}} + {\text{3CuC}}{{\text{l}}_{\text{2}}} \to {\text{2AlC}}{{\text{l}}_{\text{3}}} + 3{\text{Cu}}