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

Let total milli-mole of `Mg` used for `MgO` and `Mg_(3)N_(2)` be `a` and `b` respectively.
`{:(,2Mg+,O_(2),rarr,2MgO),("Before reaction",a,,,0),("After reaction",0,,,a),(,3Mg+,N_(2),rarr,Mg_(3)N_(2)),("Before reaction",b,,,0),("After reaction",0,,,b//3):}`
Now `(a+b//30)` milli-mole of `MgO` and `Mg_(3)N_(2)` are present in the mixture.
`MgO + 2HCl rarr MgCl_(2)+H_(2)O`,
`Mg_(3)N_(2)+8HCl rarr 3MgCl_(2) + 2NH_(4)Cl`
or the solution contains `a` milli-mole of `MgCl_(2)` from `MgO` and `b` milli-mole of `MgCl_(2)` from `Mg(3)N_(2)` and `(2b)/(3)` milli-mole of `NH_(4)Cl`.
Also millie-mole of `HCl` used for this purpose
`{:(=2a,,+,,(8b)/(3)),("for" MgO,,,,"for"Mg_(3)N_(2)):}`
Now milli-mole of `HCl` or Meq. of `HCl` (monobasic acid)
`= 60-12 = 48`
`2a + (8b)/(3) = 48`
Further, millie-mole of `NH_(4)Cl` formed `=` milli-mole of `NH_(3)` liberated
`=` milli-mole of `HCl` used for absorbing `NH_(3)`
`:. (2b)/(3) = 4` or `b = 6` ...(2)
From eqn. (1), `2a + (8 xx 6)/(3) = 48` or `a = 16`
Thus, `%` of `Mg` used for
`Mg_(3)N_(2) = (6)/((6+16)) xx 100 = 27.27%`