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

Total meq. of HCl = 60
Unreacted Meq. of HCl = meq of NaOH
Unreacted meq. of HCl = 12
Used meq. of HCl = 60 -12 = 48
Used meq. of HCl = No. of meq. of MgO and `Mg_(3) N_(2)`
`2Mg+ O_(2) rarr 2MgO`
( from air )
`MgO+ 2HCl rarr MgCl_(2) + H_(2)O`
But `Mg_(3) N_(2) + 8 HCl rarr 3 MgCl_(2) + 2NH_(4) Cl`
`NH_(4)Cl + NaOH rarr NaCl + H_(2) O + NH_(3)`
NO. of meq. of `NH_(4) Cl -= ` No. of Meq. of `NH_(3)`
`= ( 10 - 6) = 4 `
No. of meq. of `Mg_(3) N_(2) = ( 1)/( 2)` [ No. of millimole of `NH_(4) Cl] = ( 4)/( 2) = 2 `
No. of millimole ( or meq. ) of HCl used by `Mg_(3) N_(2) = 2 xx 8 = 16`
Thus, no. of meq. of HCl used by `MgO = 48 - 16 = 32 `
No. of millimoles of `MgO = ( 32)/( 2) = 16`
No. of meq. of MgO `= 16 xx 2 = 32 `
No. of meq. of Mg burtn to MgO = 32
`:.` Weight of Mg burnt to MgO `= ( 32 xx 12 )/(1000) = 0.389g `
From equation, `3Mg + N_(2) hArr Mg_(3) N_(2)`
2 millimoles of `Mg_(3) N_(2) = 6` millimoles of Mg
Weight of Mg burnt to `Mg_(3) N_(2) = ( 6 xx 24)/ ( 1000) = 0.144 g `
Total weight of Mg `= 0.384 + 0.144 = 0.528g `
% of Mg burtn of `Mg_(3) N_(2) = ( 0.144 xx 100) /( 0.528 ) = 27.27 %`