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

`MgO+2HCl rarr MgCl_(2)+H_(2)O`
`Mg_(3)N_(2)+8HCl rarr 3MgCl_(2)+2NH_(4)Cl`
`"12 meq of NaOH "-="12 meq of HCl"`
`" i.e, HCl left unreacted = 12 meq"`
`therefore" HCl used up by MgO and "Mg_(3)N_(2)=60-12="48 meq = 48 millimoles"`
Suppose in the mixture, there are x millimoles of MgO and y millimoles of `Mg_(3)N_(2)`.
Then `2x+8y=48 or x+4y=24.`
Further, `NH_(4)Cl+NaOH rarr NaCl+H_(2)O+NH_(3)`
Acid used up by `NH_(3)=10-6 = "4 meq."`
`therefore" "NH_(3)" produced = 4 meq = 4 millimoles"`
`"or "NH_(4)Cl" formed in reaction (ii) = 4 millimoles"`
This will be formed from `Mg_(3)N_(2)=2` millimoles, i.e., y = 2.
Hence,`" "x+4xx2=24 or x=16`
`" "2Mg+O_(2)rarr 2MgO`
`" "3Mg+N_(2)rarrMg_(3)N_(2)`
16 millimoles of MgO are obtained from Mg = 16 millimoles
2 millimoles of `Mg_(3)N_(2)` are obtained from Mg = 6 millimoles
Totl millimoles of Mg `=16+6=22`
Hence, Mg converted to `Mg_(3)N_(2)=(6)/(22)xx100=27.27%`