Question

What is the volume ratio of ${\text{S}}{{\text{O}}_2},{O_2}$and ${\text{S}}{{\text{O}}_3}$ in the reaction of formation of ${\text{S}}{{\text{O}}_3}$?
A. $1:2:2$
B. $2:2:1$
C. $2:1:2$
D. $1:1:2$

Answer 1

Hint: First write the equation of $S{O_2}$with ${O_2}$to form$S{O_3}$. After balancing the equation you will get the volume ratio of ${\text{S}}{{\text{O}}_2},{O_2}$and ${\text{S}}{{\text{O}}_3}$present in the formation of$S{O_3}$. The volume ratio is the same as the mole ratio.

Complete step by step answer:
In this question we have to answer the ratio of volume of compounds and products. And we know that by balancing the equations we can get the ratio of moles of the reactants with products. And also the mole ratio is the same as the volume ratio. If we find the mole ratio by balancing the equation then it is the same as the volume ratio.
The unbalanced reaction is: ${\text{S}}{{\text{O}}_2}{\text{ + }}{{\text{O}}_2}{\text{ }} \to {\text{ S}}{{\text{O}}_3}$
Now, if we want to balance it then the number of atoms on each side i.e, on the reactant side as well as on the product side must be the same. In this case, the number of sulphur atoms are the same in both sides i.e. one and then we will see the number of oxygen atoms. The number of oxygen atoms are four in reactant side and three in product side. So we have to balance the number of oxygen atoms.
So if we multiply sulphur dioxide with two then the total number of oxygen atoms in the reactant side will be six and on the product side it will be three and hence multiply sulphur trioxide with two also. Hence the final reaction after balancing the reaction will be $2S{O_2} + {O_2} \to 2S{O_3}$. In this reaction all the atoms are equal in number on both the sides i.e. number of sulphur atoms are two and number of oxygen atoms are six on both the sides.
So the volume ratio of ${\text{S}}{{\text{O}}_2},{O_2}$and $S{O_3}$ in the formation of $S{O_3}$is $2:1:2$.

So the correct option will be C i.e. $2:1:2$ .

Note:
Balance the equation correctly. And remember that the volume ratio is the same as the mole ratio. And in the formation of $S{O_3}$ from the $S{O_2}$and ${O_2}$only the product formed is $S{O_3}$ no other molecule is formed during the reaction and hence balance accordingly. And also read the sequence correctly Otherwise you may tick the wrong option.