Respuesta :
It is true that HI cannot be prepared by the reaction of Sodium iodide (NaI) and Sulphuric acid (Hâ‚‚SOâ‚„).
When solid sodium chloride (NaCl) is treated with concentrated Sulphuric acid (Hâ‚‚SOâ‚„), sodium bisulphate (NaHSOâ‚„) and hydrochloric acid (HCl) is produced.
The equation for the above reaction is as follows-
NaCl (s) + H₂SO₄ (s) → NaHSO₄ (g) + HCl (g) ↑
If, NaCl is substituted with NaI and treated with Hâ‚‚SOâ‚„, the following compounds are produced.
The equation for the above reaction is as follows-
NaI (aq) + H₂SO₄ (aq) → I₂ (s) + H₂S (g) + H₂O (l)
In the above reaction HI is not formed because Hâ‚‚SOâ‚„ is an oxidising agent and so it oxidises HI but HCl is not oxidised by Hâ‚‚SOâ‚„.
Hence, HCl is prepared by heating NaCl with conc. Hâ‚‚SOâ‚„ but HI cannot be prepared by treating NaI with conc. Hâ‚‚SOâ‚„. NaI being a reducing agent, Hâ‚‚SOâ‚„ undergoes precipitation reaction with NaI.
Learn more about Hâ‚‚SOâ‚„ here-
https://brainly.com/question/13801946
#SPJ4
