Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in a reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.

AcademicChemistryNCERTClass 10

Element M is a lustrous, divalent element so it is a metal and only very reactive metal can react with NaOH. 
 The metals Zn, Sn, and Al will react with NaOH and KOH to liberate hydrogen gas. This is because these metals are amphoteric and can react with both bases and acids. 
In the first case; a base reacts with a metal, and in the second case; an acid reacts with a metal. Hence, hydrogen gas is most likely to be produced during these reactions.

Identification of the Gas: The evolution of hydrogen gas can be checked by bringing a burning matchstick or splinter near the evolved gas. The fact that the matchstick burns with a pop sound, shows that hydrogen gas is evolved.
The reactions can be shown by the following equations:
              $2NaOH\ +\ Zn\ \rightarrow Na_{2} ZnO_{2} \ +\ H_{2} \ $
Sodium Hydroxide + Zinc (Element A) -->Sodium Zincate + Hydrogen
                $\ 2HCl\ +\ Zn\rightarrow \ ZnCl_{2} +\ H_{2} \ $
Hydrochloric Acid + Zinc --> Zinc Chloride + Hydrogen 
Updated on 10-Oct-2022 13:27:26