12 g of impure cyanogen undergoes hydrolysis by two different paths.
(i). `(CN)_(2)+4H_(2)Oto(NH_(4))_(2)C_(2)O_(4)`
(ii). `(CN)_(2)+2H_(2)OtoNH_(2)CONH_(2)`
When 11.52 g of pure ammonium carbonate `[(NH_(4))_(2)CO_(3)]` was heated, the exact amount of urea was obtained. 20 " mL of " 1.6 M acidic `KMnO_(4)` is required to completely oxidise `(NH_(4))_(2)C_(2)O_(4)`.
Q. The percentage purity of cyanogen

Options

A. `86.67%`
B. `76.67%`
C. `66.67%`
D. `56.67%`

Answer

A

Explanation

`underset(1mol)(NH_(4))_(2)CO_(3)toH_(2)N-overset(O)overset(||)underset(1mol)C-NH_(2)+2H_(2)O` (Mw of `(NH_(4))_(2)CO_(3)=96)`
Moles of urea`=(11.56)/(96)=0.12`
`MnO_(4)^(ɵ)-=(NH_(4))_(2)C_(2)O_(4)(n=2)` `[C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)]`
`20xx1.6xx5mEq-=mEq]`
`160mEq-=160mEq`
`-=(160)/(2)mmol=80mmol=0.08mol`
Therefore, moles of `(NH_(4))_(2)C_(2)O_(4)=0.08`
Let a and b " mol of "`(CN)_(2)` react in reactions (i) and (ii), respectively.
(i). `underset(a=0.08 mol)((CN)_(2)+4H_(2)O)tounderset(a=0.08mol)((NH_(4))_(2)CONH_(2))`
`thereforea=0.08,b=0.12`
Total moles of `(CN)_(2)=0.08+0.12=0.2`
Weight of `(CN)_(2)=0.2xx52=10.4`
`%` purity of `(CN)_(2)=(10.4)/(12)xx100=86.67%`