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The removal of rust from iron by treatment with tetraoxosulphate (vi) acid is…

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The removal of rust from iron by treatment with tetraoxosulphate (vi) acid is based on the

A. hydrolysis of the iron

B. reaction of acid with base

C. oxidation of the rust

D. dehydration of the iron




This is a sweet question, i call it sweet because you would need to crack your memory a bit…

Rusted iron is actually pure iron whose surface reacts with oxygen and water to form a brown coat called rust.

Now, the formula for rust is Fe2O3.H2O.

When this is treated with sulphuric acid, water and a salt called ferric sulphate is formed.

3H2SO4 + Fe2O3.H2O = 4H2O + Fe2(SO4)3

Remember ferric oxide (or rust) is amphoteric, it behaves as an acid as well as a base.

Here in the presence of sulphuric acid, it acts as a base, reacting to give water and a salt thereby removing the rust.

Now for the right answer to the above question:

  1. Option A is incorrect. hydrolysis involves addition of water, to hydrolyze iron simply implies causing it to even rust further.
  2. Option B is correct. rust is an amphoteric oxide of iron, it reacts with the acid to form salt and water, thereby removing the rust.
  3. C is incorrect. oxidation of the rust simply implies adding more oxygen to form a higher oxide Fe3O4. Sulphuric acid does not do this.
  4. D is not the correct answer. Dehydration simply involves the chemical removal of water molecule from a compound. Water cannot possibly be removed from pure iron.


You may please note these/this:

  • Option D would have been considered debatable if it featured “dehydration of the rust” and not “dehydration of the iron”.

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/ culled from 2016 JAMB-UTME Chemistry past question 25 /

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