Answer
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Hint: The metal X is a hydride of the element with an atomic number of 13 which it forms a white coloured precipitate (insoluble solid which settles at the bottom of the solution). The hydride of this element is also amphoteric because it can react with both acid and base.
Complete step by step answer:
- It is given that the metal X reacts with sodium hydroxide to give a white precipitate. From all the metal, aluminium is the metal which reacts with sodium hydroxide to give white aluminium hydroxide.
The reaction will be:
$\text{2Al + 3NaOH }\to \text{ Al(OH}{{\text{)}}_{3}}\downarrow \text{ + 3N}{{\text{a}}^{+}}$
- So, the element X is aluminium and A is aluminium hydroxide.
- But when this precipitate reacts with an excess of sodium hydroxide, it forms a complex which is soluble in the solution.
- The complex is sodium tetra hydroxy aluminate (III). The reaction is:
$\text{Al(OH}{{\text{)}}_{3}}\text{ + NaOH }\to \text{ N}{{\text{a}}^{+}}{{\left( \text{Al(OH}{{\text{)}}_{4}} \right)}^{-}}$
- So, complex B is Sodium tetra hydroxy aluminate.
- As we know that aluminium is an amphoteric so when it will react with the hydrochloric acid then it will yield aluminium chloride along with the water i.e.
$\text{Al(OH}{{\text{)}}_{3}}\text{ + 3HCl }\to \text{ AlC}{{\text{l}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{O}$
- So, compound C is aluminium chloride.
- Now, when the compound A i.e. aluminium hydroxide is heated it undergoes decomposition reaction and yields alumina (D) along with water.
$2\text{Al(OH}{{\text{)}}_{3}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\text{O}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{O}$
- From alumina, pure aluminium metal can be extracted.
Therefore, compound A, B, C, D and X are aluminium oxide, Sodium tetra hydroxo aluminate, aluminium chloride, alumina and aluminium respectively.
Note: Aluminium can be extracted from alumina with the help of electrolysis. But as alumina is a poor conductor of heat and electricity so some amount of cryolite and fluorspar is added to overcome the problem.
Complete step by step answer:
- It is given that the metal X reacts with sodium hydroxide to give a white precipitate. From all the metal, aluminium is the metal which reacts with sodium hydroxide to give white aluminium hydroxide.
The reaction will be:
$\text{2Al + 3NaOH }\to \text{ Al(OH}{{\text{)}}_{3}}\downarrow \text{ + 3N}{{\text{a}}^{+}}$
- So, the element X is aluminium and A is aluminium hydroxide.
- But when this precipitate reacts with an excess of sodium hydroxide, it forms a complex which is soluble in the solution.
- The complex is sodium tetra hydroxy aluminate (III). The reaction is:
$\text{Al(OH}{{\text{)}}_{3}}\text{ + NaOH }\to \text{ N}{{\text{a}}^{+}}{{\left( \text{Al(OH}{{\text{)}}_{4}} \right)}^{-}}$
- So, complex B is Sodium tetra hydroxy aluminate.
- As we know that aluminium is an amphoteric so when it will react with the hydrochloric acid then it will yield aluminium chloride along with the water i.e.
$\text{Al(OH}{{\text{)}}_{3}}\text{ + 3HCl }\to \text{ AlC}{{\text{l}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{O}$
- So, compound C is aluminium chloride.
- Now, when the compound A i.e. aluminium hydroxide is heated it undergoes decomposition reaction and yields alumina (D) along with water.
$2\text{Al(OH}{{\text{)}}_{3}}\text{ }\to \text{ A}{{\text{l}}_{2}}{{\text{O}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{O}$
- From alumina, pure aluminium metal can be extracted.
Therefore, compound A, B, C, D and X are aluminium oxide, Sodium tetra hydroxo aluminate, aluminium chloride, alumina and aluminium respectively.
Note: Aluminium can be extracted from alumina with the help of electrolysis. But as alumina is a poor conductor of heat and electricity so some amount of cryolite and fluorspar is added to overcome the problem.
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