Question

When zinc metal is treated with a dilute solution of a strong acid, a gas is evolved, which is utilised in the hydrogenation of oil. Name the gas evolved.
(A) carbon dioxide
(B) Hydrogen
(C) Oxygen
(D) Methane

Answer 1

Strong acids dissociate in aqueous medium to provide hydronium ions and conjugate base. Zinc acts as a reducing agent as it provides a pair of electrons and forms ${\text{Z}}{{\text{n}}^{2 + }}$ cation.

Step by step answer: When you react zinc metal with strong acids (such as hydrochloric acid or sulphuric acid) you obtain zinc salt and hydrogen gas. Write a balanced chemical equation for the reaction of zinc metal with hydrochloric acid.
${\text{Zn + 2 HCl }} \to {\text{ ZnC}}{{\text{l}}_2}{\text{ + }}{{\text{H}}_2} \uparrow$
In the above reaction, zinc metal reacts with hydrochloric acid to obtain zinc chloride and hydrogen gas. You can show the liberation of hydrogen gas by using an up-arrow $\uparrow$ after the symbol ${{\text{H}}_2}$ .
When oils are hydrogenated, they are converted to fats.
Oils contain carbon-carbon double bonds. Upon hydrogenation of oils, these carbon-carbon double bonds are converted into carbon-carbon single bonds. Each carbon-carbon double bond adds a molecule of hydrogen. Thus, you can carry out the hydrogenation of oils by using hydrogen gas in presence of metal catalysts such as platinum, palladium or nickel catalysts.
When zinc metal is treated with a dilute solution of a strong acid, hydrogen gas is evolved, which is utilised in the hydrogenation of oil. Name the gas evolved.

Hence, the correct option is the option (B).

Additional information: We can detect the liberation of the hydrogen gas by bringing a burning candle near the hydrogen gas. If you observe that the gas burns with pop sound, you can say that hydrogen gas is present.

Note: Hydronium ions are obtained when a water molecule accepts a proton. A proton accepts an electron to form a hydrogen atom. This is a reduction reaction. Two hydrogen atoms combine to give a hydrogen molecule.
${\text{2 }}{{\text{H}}^ + }{\text{ + 2 }}{{\text{e}}^ - } \to {\text{ }}{{\text{H}}_2} \uparrow$