On passing 10.0 L of a gaseous mixture of `NO_(2)` and `N_(2)` at STP, through an `NaOH` solution, a mixture of `NaNO_(2)` and `NaNO_(3)` is formed 6.32 g of `KMnO_(4)` is required to oxidise above `NaNO_(2)` in `H_(2)SO_(4)` medium. Determine the percentage by mass of gaseous mixture `(N_2` does not react with `NaOH`)

Explanation

`N_2` does not react with NaOH. So only `NO_2` reacts with NaOH.
`2NO_2+2NaOHtoNaNO_3+NaNO_2+H_2O`
`5NaNO_2+2KMnO_4+3H_2SO_4toK_2SO_4+2MnSO_4+5NaNO_3+3H_2O`
`therefore2 mol KMnO_4-=5 mol NaNO_2`
Also: `2 mol NO_2-=1 mol NaNO_2`
`implies0.1 mol NaNO_2-=0.2mol NO_2`
`-=0.2xx22.4L at STP`
`-=4.48 lit. NO_2`
or `(10-4.48)LN_2=5.52L of N_2-=0.246 " mol of "N_2`
" mol of "`NO_2=0.2-=0.2xx46=9.2g`
`% of NO_2=57.18%`
`" mol of "N_2=0.246=0.246xx28=6.28g`
`% of N_2=42.82%`