Question

Bhawna took a pale green substance A in a test tube and heated it over the flame of a burner. A brown coloured residue B was formed along with the evolution of two gases with the burning smell of sulphur. Identify A and B. Write the chemical reaction involved.

Answer 1

Hint: First make a step flowchart, to synchronize the entire process in one way; $\underset{\left( \text{pale green} \right)}{\mathop{\text{A}}}\,\xrightarrow{\vartriangle }\underset{\left( \text{brown} \right)}{\mathop{\text{B}}}\, + 2 \text{gases of sulphur}\uparrow $. The gases of sulphur would have been the oxides of sulphur. A and B are solid substances. They can be recognized by their colour.

Complete step by step answer:
Let us understand this question through the colour of the substances and properties of the product.
The brown coloured residue is a solid substance which is of brown colour. It is ferric oxide or $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}$. This compound is generally referred to as rust and rust is brown in colour. So, the compound B is Ferric oxide. It is used in steel and iron industries for the production of steel, iron and many alloys. It is clear that compound A is a compound of iron only.

The pale green of compound A is because of the presence of $\text{F}{{\text{e}}^{+2}}$ which is paramagnetic. The compound A is also known as green vitriol. The compound is sulphate of iron, which is ferrous sulphate. The chemical formula is $\text{FeS}{{\text{O}}_{4}}$. It is used in manufacturing of inks. The decomposition of $\text{FeS}{{\text{O}}_{4}}$ on heating forms ferric oxide, sulphur dioxide and sulphur trioxide. It is written as $\text{2FeS}{{\text{O}}_{4}}\left( \text{s} \right)\xrightarrow{\vartriangle }\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}\left( \text{s} \right)+\text{S}{{\text{O}}_{2}}\left( \text{g} \right)\uparrow +\text{S}{{\text{O}}_{3}}\left( \text{g} \right)\uparrow $.

$\text{S}{{\text{O}}_{2}}$ and $\text{S}{{\text{O}}_{3}}$ are the gases and oxides of sulphur having pungent smell. Sulphur trioxide is the primary agent in acid rain. Its vapours are invisible and extremely corrosive. Sulphur dioxide is a toxic gas, the odour is just like a struck match.
The compound A is ferrous sulphate and compound B is ferric oxide. The reaction is $\text{2FeS}{{\text{O}}_{4}}\left( \text{s} \right)\xrightarrow{\vartriangle }\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}\left( \text{s} \right)+\text{S}{{\text{O}}_{2}}\left( \text{g} \right)\uparrow +\text{S}{{\text{O}}_{3}}\left( \text{g} \right)\uparrow $.

Note: In this question, it is not easy to recognize the compounds. It mainly can be identified from the colour only. Generally, iron sulphates are hydrated, so, the overall reaction of this process is $\text{FeS}{{\text{O}}_{4}}.7{{\text{H}}_{2}}\text{O}\xrightarrow{\vartriangle }\text{FeS}{{\text{O}}_{4}}+7{{\text{H}}_{2}}\text{O}$. The hepta-hydrated iron sulphate loses water to form anhydrous iron sulphate. Which further on heating shows the reaction, $\text{2FeS}{{\text{O}}_{4}}\left( \text{s} \right)\xrightarrow{\vartriangle }\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}\left( \text{s} \right)+\text{S}{{\text{O}}_{2}}\left( \text{g} \right)\uparrow +\text{S}{{\text{O}}_{3}}\left( \text{g} \right)\uparrow $.