JEE Main 2020 · 9 January · evening · Single correct · +4 / −1
5 g of zinc is treated separately with an excess of
(a) dilute hydrochloric acid and
(b) aqueous sodium hydroxide.
The ratio of the volumes of H₂ evolved in these two reactions is :
- (a) 1 : 2
- (b) 1 : 1
- (c) 1 : 4
- (d) 2 : 1
Answer and solution
Answer (b) 1 : 1
Solution
Chemistry ma’am · 4 steps
Zinc loses two electrons in acid and in alkali alike, so each Zn frees one H₂ both times.
- With dilute acid. Zinc is above hydrogen in the activity series, so it displaces hydrogen from the acid: Zn + 2HCl → ZnCl₂ + H₂. Each mole of Zn gives one mole of H₂.
- With aqueous alkali. Zinc is amphoteric, so it also dissolves in excess NaOH to form sodium zincate: Zn + 2NaOH + 2H₂O → Na₂[Zn(OH)₄] + H₂. Again each mole of Zn gives one mole of H₂.
- Why they match. In both reactions Zn goes from 0 to +2 and loses 2 electrons. Two electrons always produce exactly one H₂, whether the hydrogen comes from the acid or from water. Both reagents are in excess, so the same 5 g of Zn reacts completely each time.
- So. Equal moles of Zn give equal moles of H₂. At the same temperature and pressure that means equal volumes, a ratio of 1 : 1, option (b).
The idea to keep. For gas from a metal, count electrons. The metal's change in oxidation state fixes the moles of H₂ per mole of metal, whether the reagent is an acid or an alkali.
Answer from the NTA final answer key. Solution by PrepWiser.