A sample of `Mg` was burnt in air to give a mixure of `MgO` and `Mg_(3)N_(2)`. The ash was dissolved in `60 Meq`. of `HCl` and the resulting solution was back titrated with `NaOH`. `12 Meq`. Of `NaOH` was then added and the solution distrilled. The ammonia released was then trapped in `10 Meq`. of second acid solution. Back titration of this solution required `6 Meq`. of the base Calculate the percentage of `Mg` burnt to the nitride.

Explanation

`{:(Mg+(1)/(2)O_2toMgO),(3Mg+N_2toMg_3N_2):}]`
Let x m" mol of "Mg react with `N_2` to give `Mg_3N_2` and y m mol with `O_2` to give `MgO`.
`m" mol of "MgO=y and m" mol of "Mg_3N_2=(x)/(3)`
`MgO+2"HCl"toMgCl+H_2O`
`Mg_3N_2+8"HCl"to3MgCl_2+2NH_4cl` `((2)/(3)xmmol)`
`m" mol of "HCl used=(2y+(8)/(3)x)=60-12`
`implies2y+(8)/(3)x=48`
`m" Eq of "HCl=mmole of HCl`
`12=mEq` of ecess acid
`underset((2)/(3)xmmol)(NH_4Cl)tounderset((2)/(3)x)(NH_3)`
`m" Eq of "NH_3=10-6=4`
`mmol NH_3=4` (monoacidic base)
`implies(2)/(3)x=4`
`impliesx=6mmol =` m" mol of "Mg converted to `Mg_3N_2`
Substituting in (i) and solving for y, we get
`y=16` mmol `=` m" mol of "Mg converted to MgO.
`%` Mg converted (burnt) to nitride`=((x)/(x+y))xx100`
`=(6)/(22)xx100=27.27%`