Question

Compound A is a black solid. When heated with dilute Sulphuric acid, the blue solution of B is formed. When the blue solution is treated with Sodium hydroxide solution, a blue precipitate of compound C is formed. When compound C is heated strongly, compound A is again formed. Name the compound A, B, and C.

Explain the chemical reactions and write the balanced chemical equation for the chemical changes that take place.

Answer 1

Explanation:

• Iron sulfide $\left(FeS\right)$which is black in color when heated with dilute Sulphuric acid $\left(H_2{SO}_4\right)$, the blue solution of Ferrous sulfate $\left(FeS{O}_4\right)$is formed along with Hydrogen sulfide gas $\left(H_{2}S\right)$. Further, when Ferrous sulfate reacts with Sodium hydroxide $\left(NaOH\right)$ solution, a blue precipitate of compound Iron (II) hydroxide is formed along with Sodium sulfate $\left(N{a}_{2}S{O}_4\right)$.
• The chemical reactions are as follows.

$$\begin{gathered} \underset{\mathrm{iron}\mathrm{sulfide}}{\mathrm{FeS}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Ferrous}\mathrm{sulphate}}{{\mathrm{FeSO}}_4(\mathrm{aq}})+\underset{\mathrm{Hydrogen}\mathrm{sulfide}}{{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)} \\ \underset{\mathrm{Ferrous}\mathrm{sulphate}}{{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Iron}\left(\mathrm{II}\right)\mathrm{hydroxide}}{\mathrm{Fe}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{sulfate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)} \end{gathered}$$

Thus, Compound A is Iron sulfide $\left(\mathrm{FeS}\right)$, Compound B is Ferrous sulfate$\left({\mathrm{FeSO}}_4\right)$, and compound C is Iron (II) hydroxide $\left(\mathrm{Fe}{\left(\mathrm{OH}\right)}_2\right)$.