Question

In the Baeyer’s process:
$A{{l}_{2}}{{O}_{3}}$ goes into solution as soluble $Al{{(OH)}_{4}}$ , while other basic oxides as:

a.) $Ti{{O}_{2}}$ , and $F{{e}_{2}}{{O}_{3}}$ , remain insoluble
b.) $A{{l}_{2}}{{O}_{3}}$, changes to AIN which in turn decomposed by ${{H}_{2}}O$
c.) $A{{l}_{2}}{{O}_{3}}$, changes to $A{{l}_{2}}{{(C{{O}_{3}})}_{3}}$ which changes to $AlC{{l}_{3}}$ .
d.) None of the above correct

Answer 1

To solve this question, first we have to discuss the Baeyer’s method and after that we will proceed further by understanding the process and the reactions which take place.

Complete Solution :
In this method bauxite power is used and sodium hydroxide is added to it and then it is heated in a closed vessel at 433K temperature and the pressure applied is around 5 to 6 bars for about 6 to 8 hours. The aluminium oxide present in the bauxite is converted to sodium aluminate which is soluble in water.
In bauxite other impurities like iron oxide and some other impurities are also present. Some precipitates of aluminium hydroxide are added from outside, this is done to increase the rate of the reaction. We can say that about 99.5 percent pure alumina is obtained using the Baeyer’s method. The reaction involved in the Baeyer’s process is mentioned below:

& NaAl{{O}_{2}}+2{{H}_{2}}O\to Al{{(OH)}_{3}}+NaOH \\\ & 2Al{{(OH)}_{3}}\to A{{l}_{2}}{{O}_{3}}+3{{H}_{2}}O \\\ \end{aligned}$$ Hence, the correct answer is option (A). i.e. $A{{l}_{2}}{{O}_{3}}$ goes into solution as soluble $Al{{(OH)}_{4}}$ , while other basic oxides as $Ti{{O}_{2}}$ , and $F{{e}_{2}}{{O}_{3}}$ , remain insoluble. **So, the correct answer is “Option (a)”.** **Note:** Bauxite is a rock which is formed from a reddish clay substance which is known as laterite soil. Laterite soil is found mostly in the tropical and subtropical area. Bauxite consists primarily of aluminium oxide or alumina, silica, iron oxide and many other compounds. Aluminium is used to form a variety of materials such as foils, cooking utensils etc.