Hydrogen gas is prepared in the laboratory by reacting dilute$\mathrm{HCl}$ with granulated zinc. The following reaction takes place: $\mathrm{Zn}+2\mathrm{HCl}\to {\mathrm{ZnCl}}_2+{\mathrm{H}}_2$ Calculate the volume of hydrogen gas is liberated at STP when $32.65\mathrm{g}$of Zinc reacts with$\mathrm{HCl}$ . 1 mole of a gas occupies $22.7\mathrm{L}$ volume at STP; atomic mass of $\mathrm{Zn}=65.3\mathrm{u}?$

The correct option is B

$11.35\mathrm{L}$

Explanation:

Step 1: Analyzing the given data

The given reaction; $\underset{\mathrm{Zinc}}{\mathrm{Zn}}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{2\mathrm{HCl}}\to \underset{\mathrm{Zinc}\mathrm{chloirde}}{{\mathrm{ZnCl}}_2}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2}$

Also given; Mass of $\mathrm{Zn}=32.65\mathrm{g}$

At STP, 1 mole of gas occupies = $22.7\mathrm{L}$

Atomic mass of $\mathrm{Zn}=65.3\mathrm{g}$

Step 2: Calculating the volume of hydrogen gas liberated

According to stoichiometry, 1 mole of Zinc reacts with 2 moles of hydrochloric acid to liberate 1 mole of hydrogen gas.

Now,

$65.3\mathrm{g}\mathrm{Zn}$ reacts with $\mathrm{HCl}$ to produce = $22.7\mathrm{L}\mathrm{of}{\mathrm{H}}_2$

$1\mathrm{g}\mathrm{Zn}$ reacts with $\mathrm{HCl}$ to produce = $\frac{22.7\mathrm{L}}{65.3\mathrm{g}}\mathrm{of}{\mathrm{H}}_2$

$32.65\mathrm{g}\mathrm{Zn}$ reacts with $\mathrm{HCl}$ to produce = $\frac{22.7\mathrm{L}}{63.5\mathrm{g}}\times 32.65\mathrm{g}=11.35\mathrm{L}\mathrm{of}{\mathrm{H}}_2$

Hence, option (B) is correct.