Question

10 ml of $CO$ is mixed with 25 ml air having 20 percent ${{O}_{2}}$ by volume. What would be the final volume if none $CO$ and ${{O}_{2}}$ is left after the reaction?
A.30 ml
B.40 ml
C.43 ml
D.49 ml

Answer 1

We will use Laws of Chemical Combination for Elements and Compounds. In simple terms, this law states that matter can neither be created nor destroyed. In other words, the total mass, that is, the sum of the mass of reacting mixture and the products formed remains constant.

Complete answer:
The air contains approximately 20% oxygen and 80% nitrogen.
The balanced reaction of above question is as follows:
$2CO\,+\,{{O}_{2}}\,\to \,2C{{O}_{2}}$
We are given the question that air contains 20 percent ${{O}_{2}}$ by volume.
So, 20% of 25 mL = 5 mL (oxygen in air)
Thus 25 ml of air contains 5 ml oxygen and 20 ml nitrogen.
From the reaction we get to know that
2 moles of $CO$ completely reacts with 1 mole of ${{O}_{2}}$ to give 2 moles of $C{{O}_{2}}$
1 mole of a gas = 22400 mL
2 mole of a gas = 44800 mL
So 44800 mL $CO$ will react with 22400 mL ${{O}_{2}}$ to give 44800 mL of $C{{O}_{2}}$
In the question 10 ml of co is given.
Therefore 10 mL of $CO$ will react with 5 mL ${{O}_{2}}$ to give 10 mL of $C{{O}_{2}}$
Now as there is no $CO$ and ${{O}_{2}}$ left after reaction.
Hence, the volume of the mixture after reaction is 10 ml $C{{O}_{2}}$​+ 20 ml ${{N}_{2}}$​= 30 ml.

So, the correct option is A.

Note:
For solving this type of question the balanced equation is required with the stoichiometric coefficients. Also in the above question while answering about the volume of the mixture after reaction we will have to consider nitrogen because in air apart from oxygen, nitrogen is also present.