Question

When \[10mL\] of \[{H_2}\] and \[12.5mL\] of \[C{l_2}\] are allowed to react, the final mixture contains under the same conditions.

Answer 1

Hint: Ideal gas equation: It is a law for a theoretical ideal gas. This law is the combined form of certain laws that are Boyle’s law, Charle’s law, Avogadro’s law and Gay-Lussac law. The equation is expressed as \[PV = nRT\]. For the given conditions in the question, assume the ideal behaviour of given gases.

Complete answer:
As per given reaction conditions, we need to find the final mixture under the same conditions. That means pressure and temperature are considered constant during the reaction.
So, according to the ideal gas equation: \[PV = nRT\]
Where, \[P \Rightarrow \] pressure
\[V \Rightarrow \] Volume
\[n \Rightarrow \] number of moles
\[R \Rightarrow \] Universal gas constant
\[T \Rightarrow \] temperature
Because for the given conditions, pressure and temperature are constant. Therefore, volume directly varies with the number of moles i.e., \[V \propto n\]
Now, the reaction of \[{H_2}\] and \[C{l_2}\] proceeds as follows:
\[{H_2} + C{l_2} \to 2HCl\]
As \[V \propto n\], so we can compare the ratio of reactants and products in terms of volume.
mole of \[{H_2}\] gas reacts to form \[ \Rightarrow 2\] moles of \[HCl\]
\[\therefore 10mL\] of \[{H_2}\] gas will react to form \[ \Rightarrow 10 \times 2 = 20mL\]of \[HCl\]
In the reaction, \[{H_2}\] gas is the limiting reagent and \[C{l_2}\] gas is present in excess. Therefore, the amount of \[C{l_2}\]gas left after the reaction\[ = 12.5 - 10 \Rightarrow 2.5mL\].
Hence, the final mixture contains \[20mL\] of \[HCl\] and \[2.5mL\] of \[C{l_2}\].

Note:
Limiting reagent: It is also known as limiting reactant. It is the reactant in a chemical reaction which is consumed completely after the reaction is completed and therefore, it is used to determine when the reaction will stop since the reaction cannot continue without it.