
When ferrous sulphate crystals are strongly heated, the gas/vapour not evolved are of:
[A] $S{{O}_{2}}$
[B] $S{{O}_{3}}$
[C] ${{O}_{2}}$
[D] ${{H}_{2}}O$
Answer
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Hint: Ferrous sulphate crystals contain water molecules. When we heat the crystals, the after molecules are lost. Upon further heating, we get a brown oxide compound of iron along with two different sulphur gases.
Complete step by step answer:
We know that crystals are formed due to the presence of water of crystallization. Here we have crystals of ferrous sulphate. The crystals of ferrous sulphate are light green in colour. When we heat the ferrous sulphate crystals, they turn from light green to brown. On further heating, it turns into a dark brown solid compound.
Now let us discuss the reaction of heating of ferrous sulphate crystals.
Ferrous sulphate crystals contain 7 water molecules, known as the water of crystallization. The formula is $FeS{{O}_{4}}\cdot 7{{H}_{2}}O$ . When we heat these crystals, they turn white water due to the loss of the water molecules and formation of anhydrous ferrous sulphate which is $FeS{{O}_{4}}$ thus we obtain a white coloured anhydrous compound.
\[2FeS{{O}_{4}}\cdot 7{{H}_{2}}O\xrightarrow{\Delta }2FeS{{O}_{4}}(s)+7{{H}_{2}}O(g)\]
When we further heat it, the anhydrous ferrous sulphate undergoes decomposition and forms ferric oxide along with two sulphur gases- sulphur dioxide, $S{{O}_{2}}$ and sulphur trioxide, $S{{O}_{3}}$ .
\[2FeS{{O}_{4}}(s)\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}(s)+S{{O}_{3}}(g)+S{{O}_{2}}(g)\]
As we can see from the above discussion that when we heat ferrous sulphate we obtain water in vapour form along with sulphur dioxide and sulphur trioxide.
Here, oxygen gas is not evolved.
So, the correct answer is “Option C”.
Note: In the above reaction, the single reactant which is ferrous sulphate forms three different products. It decomposes for the formation of these products therefore this reaction is a decomposition reaction. There are different kinds of decomposition reactions like photochemical decomposition, electrical decomposition or decomposition upon heating like the above reaction.
Complete step by step answer:
We know that crystals are formed due to the presence of water of crystallization. Here we have crystals of ferrous sulphate. The crystals of ferrous sulphate are light green in colour. When we heat the ferrous sulphate crystals, they turn from light green to brown. On further heating, it turns into a dark brown solid compound.
Now let us discuss the reaction of heating of ferrous sulphate crystals.
Ferrous sulphate crystals contain 7 water molecules, known as the water of crystallization. The formula is $FeS{{O}_{4}}\cdot 7{{H}_{2}}O$ . When we heat these crystals, they turn white water due to the loss of the water molecules and formation of anhydrous ferrous sulphate which is $FeS{{O}_{4}}$ thus we obtain a white coloured anhydrous compound.
\[2FeS{{O}_{4}}\cdot 7{{H}_{2}}O\xrightarrow{\Delta }2FeS{{O}_{4}}(s)+7{{H}_{2}}O(g)\]
When we further heat it, the anhydrous ferrous sulphate undergoes decomposition and forms ferric oxide along with two sulphur gases- sulphur dioxide, $S{{O}_{2}}$ and sulphur trioxide, $S{{O}_{3}}$ .
\[2FeS{{O}_{4}}(s)\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}(s)+S{{O}_{3}}(g)+S{{O}_{2}}(g)\]
As we can see from the above discussion that when we heat ferrous sulphate we obtain water in vapour form along with sulphur dioxide and sulphur trioxide.
Here, oxygen gas is not evolved.
So, the correct answer is “Option C”.
Note: In the above reaction, the single reactant which is ferrous sulphate forms three different products. It decomposes for the formation of these products therefore this reaction is a decomposition reaction. There are different kinds of decomposition reactions like photochemical decomposition, electrical decomposition or decomposition upon heating like the above reaction.
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