Question

How would calculate the mass in grams of hydrogen chloride produced when $4.9$ L of molecular hydrogen at STP reacts with an excess of chlorine gas?

Answer 1

We will calculate the moles of hydrogen by using the ideal gas equation at STP condition. Then by comparing the stoichiometry of a balanced reaction with mole of hydrogen to determine the mole of hydrogen chloride. After that we will calculate the mole of hydrogen chloride by using mole formula.

Complete step-by-step solution: The value of standard temperature in kelvin is $273\,{\text{K}}$. The value of the standard pressure in atm is ${\text{1}}\,{\text{atm}}$.
The formula of the ideal gas is as follows:
${\text{pV = nRT}}$
${\text{p}}$ is the pressure
V is the volume
${\text{n}}$ is the number of moles of ideal gas
R is the gas constant
T is the temperature
On rearranging the ideal gas equation for n we get,
${\text{n}}\,{\text{ = }}\dfrac{{{\text{pV}}}}{{{\text{RT}}}}$
The value of R in atm is ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .
On substituting ${\text{1}}\,{\text{atm}}$ for p, $4.9$ L for V, ${\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ for R and $273\,{\text{K}}$ for T,
${\text{n}}\,{\text{ = }}\dfrac{{{\text{1}}\,{\text{atm}} \times 4.9\,{\text{L}}}}{{{\text{0}}{\text{.0821}}\,{\text{Latm}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \times 273\,{\text{K}}}}$
$\Rightarrow {\text{n}}\,{\text{ = }}\dfrac{{4.9}}{{{\text{22}}{\text{.4}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}}}$
$\Rightarrow {\text{n}}\,{\text{ = }}0.218\,{\text{mol}}$
The reaction between hydrogen and chlorine is as follows:
${{\text{H}}_{\text{2}}}\, + \,{\text{C}}{{\text{l}}_{\text{2}}}\, \to \,2\,{\text{HCl}}$
According to the above reaction,
$1$ mole hydrogen = $2$ mole hydrogen chloride
$0.218$ mole hydrogen = $0.437$ mole hydrogen chloride
So, the mole of hydrogen chloride is $0.437$ mol.
Now, we will determine the mass of hydrogen chloride as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of hydrogen chloride is $36.45$ g/mol.
On substituting $0.437$ mol for moles and $36.45$ g/mol for molar mass,
$0.437\,{\text{ = }}\,\dfrac{{{\text{mass}}\,{\text{of}}\,\,{\text{HCl}}}}{{36.45}}$
${\text{mass}}\,{\text{of}}\,\,{\text{HCl}}\,\, = 0.437\, \times 36.45$
${\text{mass}}\,{\text{of}}\,\,{\text{HCl}}\,\, = 15.95$ gram
So, the mass of hydrogen chloride is $15.95$ gram.

Therefore, $15.95$ gram is the correct answer.

Note: The STP is known as standard temperature and pressure. One mole of a gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$ volume. For the stoichiometry comparison a balanced equation is necessary because without writing a balanced equation we cannot determine the actual amount of product. Molar mass of a molecule is determined by adding the atomic mass of each atom of the molecule.