Question

$5g$ of Zinc is treated separately with an excess of

I. dilute hydrochloric acid and

II. aqueous sodium hydroxide.

The ratio of the volumes of $H_2$ evolved in these two reactions is:

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $1:4$

Answer 1

The correct option is C

$1:1$

1. Reaction of zinc$\left(\mathrm{Zn}\right)$ with hydrochloric acid$\left(\mathrm{HCl}\right)$ can be represented as: $\mathrm{Zn}+\hspace{0.17em}2\mathrm{HCl}\to {\mathrm{ZnCl}}_2+{\mathrm{H}}_2$
2. Reaction of zinc$\left(\mathrm{Zn}\right)$ with sodium hydroxide$\left(NaOH\right)$ can be represented as: $\mathrm{Zn}+\hspace{0.17em}2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{ZnO}}_2+{\mathrm{H}}_2$
3. From the two reactions, it is clear that one mole of zinc gives one mole of ${\mathrm{H}}_2$in each case.
4. Hence, the ratio of the volumes of $H_2$ evolved in these two reactions$=1:1$.
5. Thus, the correct option is (C).