`SO_3` when absorbed in 98% `H_2SO_4` it forms

Kl when heated with conc . H_2SO_4 , it forms

Kl when heated with conc . H_2SO_4 , it forms

Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) Pure H_(2)SO_(4) does not react with metal because -

Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) Ionically the reaction between H_(2)SO_(4) & Mg may be presented as, Mg+2H_(3)O^(+) to Mg^(2+)+H_(2) uarr +2H_(2)O Therefore, in the given reaction -

Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) H_(2)SO_(4)+PCl_(5) to (X) overset(H_(2)O)to two strong acids where 'x' is

In the preparation of H_2SO_4 by contact process, why is SO_3 not absorbed directly in water to form H_2SO_4 ?