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Topic: Thermal decomposition of iron(II) sulphate to iron(III) oxide  (Read 45867 times)

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Ah Beng

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Hello,

I'm confused and seeking a bit of help here.


The original question :

Hydrated iron(II) sulphate is pale green. When it is heated, the crystals first turn white and on stronger heating, a brown residue of iron(III) oxide is formed. During the strong heating two gases are formed. One of the gases removes the colour from acidified potassium manganate(VII). The other gas gives a white precipitate with acidified barium chloride solution.

Construct the equations for the changes from the green crystals to the white solid, and from the white solid to the brown residue.


The (suggested) answer :

FeSO4.7H2O --> FeSO4 + 7 H2O  (reaction 1)

2 FeSO4 --> Fe2O3 + SO2 + SO3  (reaction 2)


My 1st query :

In the second reaction, as iron(II) sulphate undergoes thermal decomposition, why is iron oxidised from Fe2+ to Fe3+? If it was not given in the question that iron(III) oxide was formed, how would the candidate know that the thermal decomposition also involved a redox reaction? Do most thermal decomposition reactions involve redox?
   
   
My 2nd query :
 
Again in the second reaction, as iron(II) sulphate undergoes thermal decomposition to iron(III) oxide, is it correct to say that the first gas that is produced is sulphur dioxide, rather than sulphur trioxide?

And that some of the sulphur dioxide produced reacts (almost) immediately with oxygen in the following reversible reaction to produce sulphur trioxide :

2 SO2 + O2 --> 2 SO3  (reaction 3)

In other words, is reaction 2 a summative reaction that representatively includes reaction 3? (but this becomes problematic because oxygen is not a reactant in the left hand side of reaction 2.) Or is sulphur trioxide produced directly from the thermal decomposition of iron(II) sulphate?

Confused here, please help.


My 3nd query :

The fact that one of the gases removes the colour from acidified potassium manganate(VII), implies a reducing agent - sulphur dioxide being a well known reducing agent.

But I'm not entire sure how the other gas (sulphur trioxide) gives a white precipitate with acidified barium chloride solution.

I'm aware the barium cation (usually as barium nitrate or barium chloride) is used in ionic precipitation reaction (bonding ionically with the sulphate anion) to form the insoluble white ppt of barium sulphate; in other words, to test for the presence of sulphate anions.

However, sulphate ions aren't present in dry, covalently bonded, sulphur trioxide gas. If dissolved in water however, sulphuric acid formed, which *does* contain sulphate ions.

Is the above conjecture correct?

That the presence of sulphur dioxide is detected by its reducing capacity (seen by the decolourization of acidified potassium mangantate(VII) ), and that the presence of sulphur trioxide is detected by its forming of a sulphate ions when dissolved in water (seen by the formation of a white ppt with acidified barium chloride soln).


Thanks in advance for your clarification.

Offline Borek

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Re:Thermal decomposition of iron(II) sulphate to iron(III) oxide
« Reply #1 on: April 16, 2005, 08:26:51 AM »
The original questions lacks clarity IMHO - I thought at first that the ferric sulfate is heated with the aceces of air and I couldn't understand whole situation. But it seems it is heated without acces to air (or, more precisely, oxygen).

1. In the question final solid product was defined as Fe2O3, so the question 'how do I know what it is if it is not stated' is a philosophical rather than chemical :) But if you know that the residue was red-brown, it should be enough. FeO is black. And no- most thermal decompositions doesn't involve redox reactions. FeSO4 is a special case of reducting cation plus oxidizing anion.

2. Where are you taking O2 from? Besides, it is not necessary. Heating sulfate you decomopse it into metal oxide and sulfur trioxide. Then they react FeO + SO3 -> Fe2O3 + SO2 (unbalanced, just sketching the idea).

3. Dissolving SO3 in water you produce sulfuric acid. Period.
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