Question
Question: In the reaction; \(CO + \dfrac{1}{2}{O_2} \to C{O_2};{N_2} + {O_2} \to 2NO\) \({\text{10 mL}}\) ...
In the reaction;
CO+21O2→CO2;N2+O2→2NO
10 mL of the mixture containing carbon monoxide and nitrogen required 7 mL oxygen to form CO2 and NO, on combustion.
The volume of N2 in the mixture will be:
A) 27mL
B) 217mL
C) 4 mL
D) 7 mL
Solution
Analyze the reactions given in the question and try to relate the elements with others according to the use of the number of moles. One can also find out the relation between oxygen and nitrogen elements and relate them with the reaction of formation of nitrogen oxide and find out the correct choice.
Complete answer:
- First of all let us analyze the given reactions in the question. The reaction of formation of carbon dioxide is as,
CO+21O2→CO2
In the above reaction, one mole of carbon monoxide is reacted with a half mole of an oxygen molecule to form one mole of carbon dioxide. - Now as we know the number of moles is always directly proportional to the volume. Now let us consider the volume of N2 as x and this volume will be equal to the volume of O2 which is required to react with it.
- Now let us consider the volume of CO as y and the volume of O2 reacting with it will be 2y. By the statement of 10 mL of mixture containing carbon monoxide and nitrogen required 7 mL oxygen we can say that,
x+y=10
As 2y=7
Therefore, we can get values of x and y as 4 and 6 respectively. - We get the volume of N2 as 4 mL. And by the above reaction 21N2+21O2→NO
In the above reaction if there is 21 mole of O2 is used by CO then there will be 21 mole of O2 is used by N2 and therefore, the volume of N2 in the mixture can be 27 which shows option A as the correct choice.
Note:
The volume of the above reactions is measured in identical conditions and hence can be related to each other. The products formed in both the reactions are oxides and both reactions are carried in presence of oxygen molecules and one can relate this volume of oxygen used for the determination of the volume of nitrogen in the mixture.