Question

Question: Hydrogen gas is prepared in the laboratory by reacting dilute $HCl$ with granulated zinc. Following reaction take place:
${\text{Zn + 2HCl }} \to {\text{ ZnC}}{{\text{l}}_2}{\text{ + }}{{\text{H}}_2}$What would be the volume of the hydrogen gas is liberated at STP when $32.65{\text{ g}}$of zinc reacts with $HCl$?
A. $10.03L$
B. $11.35L$
C. $11.57L$
D. $9.53L$

Answer 1

Hint: In this question it is given that $32.65g$ of zinc reacts with dilute hydrochloric acid at STP. We also know that $1mole$ of zinc contains $65.4g$ which is the atomic mass of zinc. Now apply the simple mole concept to find the volume of hydrogen gas liberated.

Complete step by step answer:
In this question we are given a balanced equation for the reaction of zinc with dilute hydrochloric acid which generates zinc chloride and hydrogen gas.
${\text{Zn + 2HCl }} \to {\text{ ZnC}}{{\text{l}}_2}{\text{ + }}{{\text{H}}_2}$
By this balanced equation it is clear that one mole of zinc liberates one mole of hydrogen gas. We know that one mole of atom or molecule has mass equal to its molar mass and at STP has volume equal to $22.4L$. So the mass zinc in this question is given as $32.65g$ and we know from the mole concept that after balancing the equation, compare the number of moles to find the value of the required parameter. As in this question the mass of zinc is given and we have to find the volume of hydrogen gas liberated so apply mole concept as:
$65.4g$ of zinc liberates one mole i.e. $22.4L$ of hydrogen at STP.
Then $32.65g$ of zinc will liberate $\dfrac{{32.65}}{{65.4}} \times 22.4 = 11.18L$.
 Since there is no such option. So we will go through the approximation i.e. the closest value to the answer we get. So the answer will be $11.35L$

So, the correct answer is Option B.

Note:
In the question it is given that the condition is STP so we have to take the volume as $22.4L$ to calculate the volume of hydrogen gas liberated. And also apply mole concept carefully. If there is no exact answer then take the approximation i.e. mark the option which is nearest to the answer.