Question

Which of the following gas is liberated on reaction of sodium chloride with concentrated sulfuric acid?

• A. $HCl$
• B. $N{a}_{2}S{O}_4$
• C. $Both\left(A\right)and\left(B\right)$
• D. $Noneoftheabove$

Answer 1

The correct option is A $HCl$

Sodium chloride on reaction with Sulphuric acid gives hydrogen chloride gas
$NaCl+H_{2}S{O}_4\to 2HCl+N{a}_{2}S{O}_4$

Do note that here $H_{2}S{O}_4$ acts as an acid and the $HCl$ obtained is not an acid, but the Hydrogen Chloride gas, which readily escapes. On the other hand, a similar attempt with Bromide or Iodide ions would not produce their corresponding Hydrogen halides. Instead, $H_{2}S{O}_4$ would oxidize Bromide ions to $B{r}_2$.

We also know that the Iodide ion, $I^{-}$ is a much stronger reducing agent than the Bromide ion, $B{r}^{-}$. In fact, the Iodide ions are so reducing that they reduce $H_{2}S{O}_4$ all the way to $H_{2}S$ !