Question

Given, \[12\] litre of and \[11.2\] litre of are mixed and exploded .The composition of volume of mixture is
A) \[24\] Litre of \[HCl\]
B) \[0.8{\text{ }}L\] of , \[20.8{\text{ }}L\] of \[HCl\]
C) \[0.8{\text{ }}L\] and \[22.4L\] of \[HCl\]
D) \[22.4L\] of \[HCl\]

Answer 1

Hint: To attempt this question, first you will have to write a balanced chemical equation which is the prerequisite, and then since, you have moles and you have to find out the volume of product, we can easily use the ideal gas equation which is PV=nRT.

Complete answer:
Ideal gas law states that $1$ mole of an ideal gas occupies a volume of \[22.4{\text{ }}liters\].
Having this in mind, we can easily calculate the number of moles of each of the gases using their volumes given.
This can be done as follows:
If \[22.4{\text{ }}liters\] of a gas = \[1{\text{ }}mole\]
Then moles of \[12{\text{ }}liters\] of will be = \[1\; \times \dfrac{{12}}{{22.4}}{\text{ }} = {\text{ }}0.54{\text{ }}moles\]
(Note: When converting from volume to moles divide the volume by the molar volume constant which is \[22.4{\text{ }}L\] )
If \[22.4{\text{ }}L{\text{ }} = {\text{ }}1{\text{ }}mole\;\]
   \[11.2{\text{ }}L{\text{ }} = {\text{ }}1{\text{ }}mole\; \times {\text{ }}\dfrac{{11.2}}{{24}}{\text{ }} = {\text{ }}0.5\]
(Note: When converting from volume to moles divide the volume by the molar volume constant which is \[22.4{\text{ }}L\] )
Now, we have to find the limiting agent of the reaction:
The reaction between the $2$ gases is as follows:
Mole ratio is \[1:1:2\]
Therefore is the limiting reagent - as $1$ mole of requires $1$ mole of to form $2$ moles of \[HCl\] .
We use the moles of the to find the volume of the \[HCl\] .
In the reaction the mole ratio is \[1:1:2\]
Therefore, \[0.5\] moles of the will give $1$ mole of \[HCl\]
Hence, we have to calculate the volume of $1$ mole of \[HCl\]

Option C is the correct answer.

Note:
Using ideal gas law, $1$ mole of \[HCl\] will occupy \[22.4{\text{ }}liters{\text{ }}by{\text{ }}volume\] .
Hence, the total volume of the two gases
\[
   = {\text{ }}11.2{\text{ }} + {\text{ }}12{\text{ }}liters\;{\text{ }} \\
  \; = {\text{ }}\;23.2{\text{ }}liters \\
 \]
\[23.2{\text{ }} - {\text{ }}22.4{\text{ }} = {\text{ }}0.8{\text{ }}liters\] (left over or excess hydrogen).
Therefore the composition of volume of mixtures is $22.4$ litres of $HCl$ and $0.8$ litres of ${H_2}$