Question

Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride The volume of hydrogen gas produced at STP from the reaction of $115 \, g$ of zine with excess $HCl$ is ___$L$ (Nearest integer)
(Given: Molar mass of $Zn$ is $654 \,g \,mol ^{-1}$ and Molar volume of $H _2$ at $STP =227 \,L$ )

Hint:

To calculate the volume of gas produced in reactions at STP, use the molar volume of the gas (22.7 L/mol for H\(_2\) at STP).

Answer 1

Correct Answer: 4

The correct answer is 4.
$Zn+2HCl\to ZnC{l}_2+H_2\uparrow$
$\text{Moles of}Zn\text{used}=\frac{11.5}{65.4}=\text{Moles of}{H}_2\text{evolved}$
$\text{Volume of}{H}_2=\frac{11.5}{65.4}\times 22.7L=3.99L$
$=4$

Answer 2

The balanced chemical equation for the reaction is:

\[ \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \uparrow \]

The moles of Zn used is:

\[ \text{Moles of Zn} = \frac{11.5 \, \text{g}}{65.4 \, \text{g/mol}} = 0.176 \, \text{mol} \]

Since the mole ratio between Zn and H\(_2\) is 1:1, the moles of H\(_2\) produced is also 0.176 mol. Using the molar volume of H\(_2\) at STP, the volume of H\(_2\) is:

\[ \text{Volume of H}_2 = 0.176 \, \text{mol} \times 22.7 \, \text{L/mol} = 3.99 \, \text{L} \] = 4 L

This calculation assumes that the hydrogen gas is at standard temperature and pressure (STP), where one mole of gas occupies 22.7 liters. The result shows the volume of hydrogen gas produced from the reaction of zinc with hydrochloric acid.