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 followi

1.	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 -
Answer» oxidising AGENT is `H^(+)(H_(2)O)` reducingagent is `H^(+)(H_(2)O)` oxidising agent is `SO_(4)^(2-)` oxidising agent are `SO_(4)^(2-)` & `H_(3)O^(+)` both Solution :`H_(2)SO_(4)` act as oxidizing agent. `H_(2)SO_(4)+2PCl_(5) to UNDERSET((X))(SO_(2)Cl_(2))+2POCl_(3)+2HCl`