How many litre of Cl_2 at STP \:will\: be\: liberatedSTP will\: be\: liberated by the oxidation of NaCl with 10g KMnO_4 ?

  • Option 1)

    3.54l

  • Option 2)

    7.08l

  • Option 3)

    1.77l

  • Option 4)

    None of these

 

Answers (1)
A Aadil Khan

As we learnt in 

Oxidation -

It is defined as the addition of oxygen / electronegative element to a substance or removal of hydrogen / electropositive element from a substance.

- wherein

e.g. 2H_{2}S(g) + O_{2}(g)\rightarrow 2S(s)+2H_{2}O(l)

 

 

Reduction -

Removal of Oxygen / electronegative element from a substance of addition of hydrogen / electropositive element to substance.

- wherein

e.g.2HgO\overset{\bigtriangleup }{\rightarrow}2Hg(l) + O_{2}(g)

 

The reaction undergoes as follows

10\, \, NaCl+2KMnO_{4}\rightarrow SCl_{2}+\, \, side\, \, products

\therefore 2\times 158g\, \, of\, \, KMnO_{4}\rightarrow 5\times 22.4L\, \, at\, \, STP

316g\, \, of\, \, KMnO_{4}\rightarrow 5\times 22.4L\, \,

10g\, \, of\, \, KMnO_{4}\rightarrow \frac{5\times 22.4\times 10L}{316}=3.5L

\therefore Solution\, \, is\, \, 1


Option 1)

3.54l

Correct

Option 2)

7.08l

Incorrect

Option 3)

1.77l

Incorrect

Option 4)

None of these

Incorrect

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