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 $S{O_2}$would be required?
a.) 12.8 g
b.) 6.4 g
c.) 0.2 mole
d.) 0.1 mole

Answer 1

Hint: The reaction that occurs between $S{O_2}$and $CaS{O_3}$ which result in disappearing of milkiness of $CaS{O_3}$ can be written as-
$S{O_2} + CaS{O_3} + {H_2}O \to Ca{(HS{O_3})_2}$
The reaction clears us that one mole of $CaS{O_3}$ requires one mole of $S{O_2}$ gas to be passed through it. Thus, by changing the terms in the number of moles, we can find out.

Complete step by step answer:
First, let us see the reaction that is occurring between $S{O_2}$and $CaS{O_3}$. This can be written as-
$S{O_2} + CaS{O_3} + {H_2}O \to Ca{(HS{O_3})_2}$
It has been told in question that on passing $S{O_2}$through $CaS{O_3}$, the milkiness disappears.
Further, let us write the values given to us.
Given :
Mass of $CaS{O_3}$= 12 g
What we need to find :
Amount of $S{O_2}$required for 12 g of $CaS{O_3}$.
Now, let us begin.
If we see in terms of moles, then from the balanced equation above; we see that one mole of $CaS{O_3}$ reacts with one mole of $S{O_2}$ only.
Thus, if we calculate the amount of $CaS{O_3}$ in terms of moles; then it would be easier to find out.
Number of moles of $CaS{O_3}$ = $\dfrac{{Mass{\text{ of solute given i}}{\text{.e}}{\text{. }}CaS{O_3}}}{{Molar{\text{ mass of }}CaS{O_3}}}$
Thus, Number of moles of $CaS{O_3}$ = $\dfrac{{12g}}{{120g}}$
Number of moles of $CaS{O_3}$ = 0.1 mole
So, as we saw that one mole of $CaS{O_3}$ reacts with one mole of $S{O_2}$ only. So, the 0.1 moles of $CaS{O_3}$will react with 0.1 moles of $S{O_2}$ only.

Thus, the correct answer is 0.1 mole.
So, the option d.) is the correct answer.

As the mass is given in terms of grams also. So, let us convert the 0.1 moles of $S{O_2}$into grams as-
Mass of $S{O_2}$= Number of moles of $S{O_2}$$ \times $ Molar mass of $S{O_2}$
Mass of $S{O_2}$= 0.1$ \times $64
Mass of $S{O_2}$= 6.4 g

So, even the option b.) is also correct.
Thus, the correct answer is option b.) and option d.).

Note: The calcium sulphite is milky white initially. When we pass the sulphur dioxide through it, it results in the disappearance of whiteness resulting in a clear solution of the product.
Further, it is always easy to work in terms of moles because the balanced chemical reaction is represented in terms of moles only.