Question

$ 12\text{ }L $ of $ {{H}_{2}} $ and $ 11.2\text{ }L $ of $ C{{l}_{2}} $ are mixed and exploded. The composition by volume of mixture is:
(A) $ 24\text{ }L $ of $ HC{{l}_{\left( g \right)}} $
(B) $ 0.8\text{ }L\text{ }C{{l}_{2}} $ and $ 20.8\text{ }L\text{ }HC{{l}_{\left( g \right)}} $
(C) $ 0.8\text{ }L\text{ }{{H}_{2}} $ and $ 22.4\text{ }L\text{ }HC{{l}_{\left( g \right)}} $
(D) $ 22.4\text{ }L\text{ }HC{{l}_{\left( g \right)}} $

Answer 1

Hint :We know that the given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent. The limiting reagent gives the moles of product formed in the reaction.

Complete Step By Step Answer:
As we know, the quantitative study of the reactants and products involved in a chemical reaction is known as Stoichiometry, which is concerned with numbers. This concept helps to balance the chemical reactions to calculate the number of reactants and products. Generally, all the reactions depend on the substance required for the reaction. Always remember that the limiting reagent is that reactant which will be completely consumed first during the course of a reaction. Also remember that the contraction in the volume is basically the term which is defined as the decrease in the volume of that particular substance.
The balanced equation for this reaction is: $ {{H}_{2}}+C{{l}_{2}}\to 2HCl $
Thus, one mole of $ {{H}_{2}} $ ​ reacts with one mole of $ C{{l}_{2}} $ to give two moles of $ HCl. $ Since $ 22.4 $ litres of gas occupies one mole. Moles of $ 12\text{ }L $ of $ {{H}_{2}} $ will be $ =\dfrac{12}{22.4}=0.54 $ moles similarly moles of $ 11.2\text{ }L $ of $ C{{l}_{2}} $ gas $ =\dfrac{11.2}{22.4}=0.5 $ moles.
 $ C{{l}_{2}} $ is the limiting reagent; it is present in a limited amount. Therefore, $ 0.5 $ moles of the $ C{{l}_{2}} $ ​ will give one mole of $ HCl. $ Since $ {{H}_{2}} $ was in excess, hence some part of it is unreacted.
So moles of unreacted $ {{H}_{2}} $ ​ is $ \dfrac{12}{22.4}-\dfrac{11.2}{22.4}=0.035 $ moles and Volume of unreacted $ {{H}_{2}} $ ​ is $ =0.035\times 22.4=0.8 $ litres.
Therefore, the composition of the volume of mixtures is $ 22.4 $ litres of $ HCl $ and $ 0.8 $ litres of $ {{H}_{2}}. $
Therefore, the correct answer is option C.

Additional Information:
The number of molecules that participate in a reaction is called a Stoichiometric number, which represents a number in front of atoms or molecules or ions in a reaction. This number may be a fraction or number. With the help of stoichiometry, can determine how much substance for concern reaction is needed or present. Reactants and product mass, molecular weight, chemical equations and formulas are able to be measured by this Stoichiometry concept. In a given useful equation, the desired amount of products will produce from the number of materials determined by using stoichiometry in industrial applications

Note :
Remember that the limiting reagent in a reaction is the species supplied in an amount smaller than that required by the stoichiometric relation between the reactants. In the above reaction the limiting reagent is chlorine. Always write a balanced equation in the questions of stoichiometry.