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Question: 1L of \[C{{O}_{2}}\] is passed over red hot cake. The volume becomes 1.4L, the composition of the pr...

1L of CO2C{{O}_{2}} is passed over red hot cake. The volume becomes 1.4L, the composition of the product (s) is:
(A) 0.6 L CO
(B) 0.8 L CO2C{{O}_{2}}
(C) 0.6 L CO2C{{O}_{2}} and 0.8 L CO
(D) 0.8 L CO2C{{O}_{2}} and 0.6 L CO

Explanation

Solution

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.

Complete step by step answer:
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.
Given that red hot cake reduces CO2C{{O}_{2}} to CO.
The balanced equation is,
CO2+C2COC{{O}_{2}}+C\to 2CO

Therefore from the above equation, 1 mole of CO2C{{O}_{2}} will give 2 moles of CO
In the given question, 1L CO2C{{O}_{2}} will react to give 1.4 L of the mixture.
Out of 1L, if x L of CO2C{{O}_{2}} reacts, then the quantity which is not involved in reaction 1-x.
The total volume of the mixture = unreacted CO2C{{O}_{2}} + volume of CO formed
Given, total volume of the mixture = 1.4 L = 1 - x + 2x
Hence, x = 0.4L
Thus, volume of CO formed = 2x = 2X0.4L2 X 0.4L =0.8 L
Volume of CO2C{{O}_{2}} left = 1 - x = 1 - 0.4 = 0.6 L
Therefore, the compositions of the products are 0.6 L CO2C{{O}_{2}} and 0.8 L CO.
So, the correct answer is “Option C”.

Note: 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 able to measure 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.