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

First Method : m.e method
m.e of MgO + m.e of `Mg_(3)N_(2)` = m.e of HCl reacted
= m.e of total hCl - m.e of NaOH
` = 60 - 12 = 48`
In the dissolution of ash , HCl reacts with total Mg in `Mg_(3)N_(2)` and in MgO and also with N in `Mg_(3)N_(2)`
` :. ` m.e of total Mg + m.e of N = 48
or m.e of total mg + m.e of `NH_(3) = 48`
m.e of total Mg = `48 - 4 - 44`
Further , Mg converted to `Mg_(3)N_(2)` whose N converts to `NH_(4)Cl ( or NH_(3))`
` :. ` m.e of Mg converted to `Mg_(3)N_(2) = 3xx m.e of NH_(3)`
` = 3xx (10-6)`
= 12
` :. ` percentage of Mg converted to `Mg_(3)N_(2) = 12/44 xx 100`
`= 27.27 % `
Second Method : Mole method
The reactions involved are
The reactions involved are `{:(Mg to,MgO,MgO,+ 2HCl = MgCl_(2)+H_(2)O),(" x mmol(say)","x mmol"," x mmol",):}`
`{:(Mg to,Mg_(3)N_(2),Mg_(3)N_(2)+ 8HCl ,= 3MgCl_(2),2NH_(4)Cl),(" y mmol (say)",y/3mmol,y/3mmol,,(2y)/3mmol):}`
` :. 2x ` mmol of `hCl + (8y)/3` mmol of HCl
= total mmol of HCl - mmol of NaOH
`= 60 - 12 = 48`
`2x + (8y)/3 = 48 " "...(1)`
Further ,mmol of `NH_(4)Cl` = mmol of `NH_(3) = (10-60)`
or `(2y)/3 = 4 " " ...(2)`
From eqns . (1) and (2) one can calculate x = 16 and y = 6
` :. ` percentage of Mg converted to `Mg_(3)N_(2) = y/(x+y) = 27.27 % `