Question

8g of sulphur is burnt to form $S{{O}_{2}}$ which is oxidised by $C{{l}_{2}}$ water. The solution is treated with $BaC{{l}_{2}}$ solution. Moles of $BaS{{O}_{4}}$ precipitated are:
(A)- 1.0 mole
(B)- 0.5 mole
(C)- 0.75 mole
(D)- 0.25 mole

Answer 1

Hint: From the Law of conservation of mass, the mass remains conserved throughout the reaction process. So, the moles in the reaction remains conserved. Also, the precipitate is formed by the double displacement reaction.

Complete answer:
When 8 g of sulfur is burnt, it combines with the oxygen present in the air to form sulphur dioxide. So, a redox reaction takes place as follows:
$S(s)+{{O}_{2}}(g)\to S{{O}_{2}}(aq)$ - (a)
-The sulphur dioxide produced is further reacted with the chlorine and water, giving the reaction as follows:
$S{{O}_{2}}+C{{l}_{2}}+2{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}+2HCl$ - (b)
In this reaction, both the sulphur dioxide and the chlorine are acidic in nature as they react with water forming strong acids, that is sulphuric acid and hydrochloric acid.
-The above reaction is followed by the treatment of the solution with barium chloride, which takes place as follows:
$BaC{{l}_{2}}(aq)+{{H}_{2}}S{{O}_{4}}(aq)\to BaS{{O}_{4}}(s)+2HCl(aq)$ - (c)
This is a double- displacement reaction, with the formation of the barium sulphate precipitate during the reaction. So, as given the weight of sulfur, the moles will be $=\dfrac{weight}{molar\,mass}=\dfrac{8}{32}=0.25\,moles$
As during the complete process, involving the reaction (a), (b) and (c), the initial mass in the reaction remains conserved throughout.

So, the moles of the barium sulphate precipitate are option (D)- 0.25 moles.

Note: In both the redox reactions, in reaction (a) the sulfur changes in its oxidation state from 0 to (+4), acting as a reducing agent and oxygen from 0 to (-2), acting as a reducing agent. And in reaction (b), the oxidation state of sulfur changes form (+4) to (+6) and chlorine from 0 to (-1).