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How will you isolate CH(4) , C(2) H(4) a...

How will you isolate `CH_(4) , C_(2) H_(4) and C_(2) H_(2)` from their mixtures ?

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1 The mixture of methane `(CH_(4))` , ethylene `(C_(2)H_(4))` and acetylene `(C_(2)H_(2))` gases is at first passed through ammoniacal `Cu_(2) Cl_(2)` solution . Acetylene is absorbed in ammoniacal `Cu_(2)Cl_(2)` solution with the formation of red precipitate of cuprous acetylide . Methane and ethylene pass out without undergoing any reaction .
`HC -= CH + Cu_(2) Cl_(2) overset(NH_(4)OH)(to) CuC-= C Cu darr ` (red)
2 The red precipitate is filtered , washed with alcohol and then boiled with concentrated HCl or KCN solution when acetylene gas is evolved . It is dried by `P_(2)O_(5)` and collected .
`CuC-= C Cu + 2 HCl to HC -= CH uarr + Cu_(2) Cl_(2)`
3 The gas mixture (`CH_(4)` and `C_(2)H_(4)` ) which escapes is then passed through fuming sulphuric acid . `CH_(4)` comes out without any reaction . It is collected after removing acid vapours by passing through KOH . Ethylene reacts with fuming sulphuric acid to form ethyl hydrogen sulphate which when heated at `170^(@)C` liberates ethylene gas . It is collected after removing acid vapours by passing through KOH .
`H_(2) C = CH_(2) + H_(2)SO_(4) to CH_(3) CH_(2) OSO_(3) H CH_(3) CH_(2) OSO_(3) H overeset( 170^(@) C)(to) CH_(2) = CH_(2) uarr + H_(2)SO_(4)`
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