A mixture of `N_(2) NO` and `NO_(2)` was analysed by selctive absorption of the oxides of nitrogen The initial volume of the mixture was `2.74 cm^(3)` After treatment with water which absorbed the `NO_(2)` the volume was `2.02 cm^(3)` A ferrous sulphate solution was then shaken with the residual gaseas after which the volume was `0.25 cm^(3)` All volumes were measured at same `P` and `T` neglecting water vapour what was the volume % of each gas in the origional mixture ? .

Explanation

`V_(N_(2)) + V_(NO) + V_(NO_(2)) =2.74 cm^(3)`
`V_(N_(2)) +V_(NO) = 2.02 cm^(3)` (water absorbs `NO_(2))`
`V_(N_(2)) = 0.25 cm^(3) (FeSO_(4)) absorbs`NO`
`V_(NO) = 1.77 cm^(3)`
`V_(NO_(2)) = 0.72 cm^(3)`
`:. % N_(2) = (0.25)/(2.74) xx 100 = 9.12%`
`% NO = (1.77)/(274) xx 100 = 64.60%`
`% NO_(2) = (0.72)/(2.74) xx 100 = 26.28%` .