Saved Bookmarks
| 1. |
A sampleof Mgwas burnt in air to give a mixture of MgO and Mg_(3)N_(2).The ash was disolvedin 60 m.e of HCl and the resulting solution back titrated with NaOH . 12 m.e of Naoh were required to reach the end point . An excess of NaOH was then added and the solution distilled .the ammonia released was then irapped in 10 m.e of second acid solution . Back titration of this solution required 6 m.e of the base . Calculate the percentage of Mg burnt to nitrate . |
|
Answer» Solution :First Method : m.e method m.e of MgO + m.e of `Mg_(3)N_(2)` = m.e of HCL reacted = m.e of totalhCl - 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 cancalculate x = 16 and y = 6 ` :. ` percentage of Mgconvertedto `Mg_(3)N_(2) = y/(x+y) = 27.27 % ` |
|