Question

1. A black mineral (A) on heating in the presence of air gives a gas (B).
2. The mineral (A) on reaction with dilute ${{H}_{2}}S{{O}_{4}}$ gives a gas (C) and solution of a compound (D).
3. On passing gas (C) into an aqueous solution (B), a white turbidity is obtained.
4. The aqueous solution of compound (D) on reaction with potassium ferricyanide gives a blue compound (E).
Compounds (A) to (E) are identified as:
(A) $FeS$
(B) $S{{O}_{2}}$
(C) ${{H}_{2}}S$
(D) $~FeS{{O}_{4}}$
(E) $K\overset{II}{\mathop{Fe}}\,\left[ \overset{III}{\mathop{Fe}}\,{{\left( CN \right)}_{6}} \right]$

Answer 1

Hint : We know that the displacement reaction is the one in which the atom, or a set of atoms, is displaced by another atom or set of atoms, in a molecule. Exothermic reactions are reactions that release energy, generally in the form of heat or light.

Complete Step By Step Answer:
Sulfur dioxide or sulphur dioxide is the chemical compound with the formula $S{{O}_{2}}.$ It is a toxic gas responsible for the smell of burnt matches. It is released naturally by volcanic activity and is produced as a byproduct of copper extraction and the burning of fossil fuels contaminated with sulfur compounds. The Claus process is the most significant gas desulfurization process, recovering elemental sulfur from gaseous hydrogen sulfide. Hydrogen sulfide produced, for example, in the hydro-desulfurization of refinery naphtha and other petroleum oils, is converted to sulfur in Claus plants. The black mineral (A) is ferrous sulphide, $FeS.$ The gas obtained from heating in the presence of air is sulphur dioxide $S{{O}_{2}}.$
$2FeS+\dfrac{7}{2}{{O}_{2}}\to F{{e}_{2}}{{O}_{3}}+2S{{O}_{2}};~~~~FeS$ reacts with dil sulphuric acid to form hydrogen sulphide gas and the compound (D) which is ferrous sulphate, $FeS{{O}_{4}}.$ $FeS+{{H}_{2}}S{{O}_{4}}\to FeS{{O}_{4}}+{{H}_{2}}S;~~~~FeS$ reacts with gas (C) which is ${{H}_{2}}S~$ and gives white turbidity of $\,S.$
$S{{O}_{2}}+2{{H}_{2}}S\to 2{{H}_{2}}O+3S$ Ferrous sulphate reacts with potassium ferricyanide to form a blue compound (E), which is $K\overset{II}{\mathop{Fe}}\,\left[ \overset{III}{\mathop{Fe}}\,{{\left( CN \right)}_{6}} \right]$ ; $FeS{{O}_{4}}+{{K}_{3}}[Fe{{(CN)}_{6}}]\to K\overset{II}{\mathop{Fe}}\,\left[ \overset{III}{\mathop{Fe}}\,{{\left( CN \right)}_{6}} \right].$
Therefore, the correct answer is option C.

Note :
Remember that the conventional Claus process described above is limited in its conversion due to the reaction equilibrium being reached. Like all exothermic reactions, greater conversion can be achieved at lower temperatures, however as mentioned the Claus reactor must be operated above the sulfur dew point.