Question

In the formation of $S{O_2}$ and $S{O_3}$ ​, the ratio of the weights of oxygen, which combines with $10\,\,kg$ of sulphur is:
A) $1:1$
B) $3:2$
C) $2:3$
D) $3:4$

Answer 1

To solve this question, we must first understand some basic concepts of Mole concept. Then we need to use the basic concepts and logic for calculation of the required ratio and then only we can conclude the correct answer.

Complete solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
Law of multiple proportions: states that if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers. This law is sometimes called Dalton's Law, named after John Dalton, the chemist who first expressed it.
Step 1: In this step we will enlist first reaction involved in this:
$S\,\, + \,\,{O_2}\,\, \to \,\,S{O_2}$
Here in the reaction, $32\,\,kg$ of Sulphur will react with $32\,\,kg$ of oxygen gas.
And therefore, $10\,\,kg$ of Sulphur will react with $10\,\,kg$ of oxygen gas.
Step 2: In this step we will enlist second reaction involved in this:
$2S\,\, + \,\,3{O_2}\,\, \to \,\,2S{O_3}$
Clearly in the above reaction, $64\,\,kg$ of Sulphur will react with $96\,\,kg$ of oxygen gas.
And therefore, $10\,\,kg$ of Sulphur will react with $15\,\,kg$ of oxygen gas.
Step 3: Ratio calculation:
Ratio of oxygen required $= \,10:15\,\, = \,\,2:3$
Hence the required ratio is $2:3$

Hence the correct answer is C.

Note: The reaction of $S{O_2}$ with fly ash in the presence of ${O_2}$ and ${H_2}O$ involves a series of reactions that lead to the formation of SO3 and eventually ${H_2}S{O_4}$ . Homogeneous experiments were conducted to evaluate the effects of the procedural variables, i.e., temperature, gas concentrations, and residence time, on the post-combustion conversion of $S{O_2}$ to $S{O_3}$ .