Question

The volume ratio of $S{O_2},\,{O_2}$ and $S{O_3}$ in the reaction for the formation of $S{O_3}$ is:
(i) $1:2:2$
(ii) $2:2:1$
(iii) $2:1:2$
(iv) $1:1:2$

Answer 1

For this question we know which the reactants are and which the product is so we need to at first write the equation corresponding to the reaction. Then balance the reaction using suitable steps and obtain a balanced equation for the reaction then find the volume ratio from the balanced equation.

Complete step-by-step answer: It is given that by the reaction $S{O_3}$ is formed hence $S{O_2}$ and ${O_2}$ are the reactants of the said reaction. So the reaction proceeds as: $S{O_2} + {O_2} \to S{O_3}$.
Now, we need to balance the equation for finding the volume ratio in which they react.
To balance the equation we need to count all of the atoms on each side of the chemical equation.
As we can see we have $1\,S$ atom on the left hand side of the equation, hence we need $1\,S$ atom on the right side of the equation too. Hence the equation becomes: $1S{O_2} + {O_2} \to 1S{O_3}$.
Now, we have $3\,O$ atoms on the right, hence we need $3\,O$ atoms on the left side too. But there are already $4\,O$ atoms on the left so we must put a $\dfrac{1}{2}$ in front of ${O_2}$ on the left. So now the equation becomes: $1S{O_2} + \dfrac{1}{2}{O_2} \to 1S{O_3}$.
Now, multiplying the whole equation by $2$ we get: $2S{O_2} + 1{O_2} \to 2S{O_3}$ which is the balanced equation.
The reactants and products react in the volume ratio as depicted by the stoichiometry in the balanced equation. Hence the volume ratio of $S{O_2},\,{O_2}$and $S{O_3}$ is $2:1:2$.

So the correct answer is (iii) $2:1:2$.

Note: For this question you must balance the equation properly. Only change the numbers in front of compounds i.e. the coefficients. Never change the numbers after atoms i.e. the subscripts of the atoms. The number of each atom on both sides of the equation must be the same for the balanced equation.