In the reaction given
`Mg(s)+NO_(3)^(ɵ)+H_(2)OtoMg(OH)_(2)(s)+overset(ɵ)(O)HO(aq.)+NH_(3)(g)` 20 " mL of " sample of `NO_(3)^(ɵ)` solution is treated with Mg. The `NH_(3)^(g)` was passed into 50 " mL of " 0.1 M HCl. The excess HCl requried 30 " mL of " 0.1 M KOH for its neutralisation calculate the molarity of `NO_(3)^(ɵ)` ions in the original sample?

Explanation

Balance the equation in basic medium
`cancel(8e^(-))+6H_(2)O+NO_(3)^(ɵ)toNH_(3)+9overset(ɵ)(O)H`
`underline(2(ɵ)(O)H+"Mg"to overset(+2)(M)g(OH)_(2)+cancel(2e^(-))]xx4)`
`underline(6H_(2)O+NO_(3)^(ɵ)+4"Mg"toNH_(3)+4Mg+4Mg(OH)_(2)+overset(ɵ)(O)H)`
`thereforem" Eq of "NH_(3) "formed"=m" Eq of "HCl" used for "NH_(3)`
`=(50xx0.1xx1)-(30xx0.1xx1)`
`=2 mEq`
Thus, m" Eq of "`NH_(3)` for valence factor of `8=8xx2=16`
Also, mEw of `NO_(3)^(ɵ)=m" Eq of "NH_(3)`
`=8xx2=16mEq`
`N_(NO_(3)^(ɵ))xxV_(mL)=16`
`therefore N_(NO_(3)^(ɵ))xx20=16`
`N_(NO_(3)^(ɵ))=(16)/(20)=0.8M`
`M_(NO_(3)^(ɵ))=(0.8)M`
`M_(NO_(3)^(ɵ))=(0.8)/(8)=0.1M`