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Question: When \[10mL\] of \[{H_2}\] and \[12.5mL\] of \[C{l_2}\] are allowed to react, the final mixture cont...

When 10mL10mL of H2{H_2} and 12.5mL12.5mL of Cl2C{l_2} are allowed to react, the final mixture contains under the same conditions.

Explanation

Solution

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=nRTPV = 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=nRTPV = nRT
Where, PP \Rightarrow pressure
VV \Rightarrow Volume
nn \Rightarrow number of moles
RR \Rightarrow Universal gas constant
TT \Rightarrow temperature
Because for the given conditions, pressure and temperature are constant. Therefore, volume directly varies with the number of moles i.e., VnV \propto n
Now, the reaction of H2{H_2} and Cl2C{l_2} proceeds as follows:
H2+Cl22HCl{H_2} + C{l_2} \to 2HCl
As VnV \propto n, so we can compare the ratio of reactants and products in terms of volume.
mole of H2{H_2} gas reacts to form 2 \Rightarrow 2 moles of HClHCl
10mL\therefore 10mL of H2{H_2} gas will react to form 10×2=20mL \Rightarrow 10 \times 2 = 20mLof HClHCl
In the reaction, H2{H_2} gas is the limiting reagent and Cl2C{l_2} gas is present in excess. Therefore, the amount of Cl2C{l_2}gas left after the reaction=12.5102.5mL = 12.5 - 10 \Rightarrow 2.5mL.
Hence, the final mixture contains 20mL20mL of HClHCl and 2.5mL2.5mL of Cl2C{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.