Question

$S{O_2}$ gas is slowly passed through an aqueous suspension containing 12 g $CaS{O_3}$ till the milkiness just disappears. What amount of $S{O_2}$ when required?
A.6.4 mole
B.0.3 mole
C.0.2 mole
D.0.1 mole

Answer 1

Hint: The uptake of gas phase $S{O_2}$ by aqueous solutions was studied as a function of pH and temperature using droplet train reactor apparatuses. All atmospheric species react with water and $OH^-$, acidifying aqueous solutions. This reaction is the dependence on the reaction of desulfurization reagent. Here this is an acid-base reaction where CaO is a base, and $S{O_2}$ is an acid; white or grey, odorless lumps or granulate powder.

Complete step by step answer:
Here calcium sulphate reacts with the process of Sulphur dioxide and water and forms the product. The reaction is given below:
 $CaS{O_3} + S{O_2} + {H_2}O \to {(HS{O_3})_2}Ca$
Mole of $CaS{O_3} = \dfrac{{12}}{{120}} = 0.1$ mole
1 mole $CaS{O_3}$ = 1 mole $S{O_2}$
Here we calculate for 0.1 mole,
mole $CaS{O_3} = \dfrac{{0.1 \times 1}}{1}$
 =0.1 mole of $S{O_2}$
Here we find out the mass of $S{O_2}$ which is given below,
Mass of $S{O_2} = 0.1 \times 64$
= 6.4 gm = 6.4 mole

Hence option (A) is the correct answer.

Additional information:
Sulphur dioxide irritates the skin and mucous membranes of the eyes, nose, throats and lungs. High concentration of Sulphur dioxide can cause inflammation and irritation of the respiratory system, especially during heavy physical activity.

Note:Here we see that the calculation of 1 mole of $CaS{O_3}$ is similar to the 1 mole of $S{O_2}$. As well as we take 0.1 mole of $CaS{O_3}$ is also similar to 0.1 mole of $S{O_2}$. Mass is found by the use of given moles. Here we remember that the moles of compounds are the same.