The half time of first order decomposition of nitramide is `2.1` hour at `15^(@)C`.
`NH_(2)NO_(2(aq.))rarr N_(2)O_((g))+H_(2)O_((l))`
If `6.2 g` of `NH_(2)NO_(2)` is allowed to decompose, calculate:
(i) Time taken for `NH_(2)NO_(2)` is decompose `99%`.
(ii) Volume of dry `N_(2)O` produced at this point measured at STP.

Explanation

6,2 gm of `NH_(2)NO_(2)=0.1` mol and 1 mole `NH_(2)NO_(2)=1` mole of `N_(2)O` As 99% of `NH_(2)NO_(2)` is decomposed `rArr 0.099` mol of `NH_(2)NO_(2)` is decomposed `0.099` mol of `N_(2)O` are produced `equiv 22.4xx0.099`
`=2.217 L" of "N_(2)O" at STP."`