A sample of magnesium was burnt in air to give a mixture of MgO and Mg_(3)N_(2). The ash was dissolved in 60 meq of HCl and the resulting solution back titrated with NaOH. 12 meq 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 trapped in 10 meq of second acid solution. Back titration of this solution required 6 meq of the base. Calculate the percentage of magnesium burnt to the nitride.

A sample of magnesium was burnt in air to give a mixture of MgO and Mg

1.	A sample of magnesium was burnt in air to give a mixture of MgO and Mg_(3)N_(2). The ash was dissolved in 60 meq of HCl and the resulting solution back titrated with NaOH. 12 meq 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 trapped in 10 meq of second acid solution. Back titration of this solution required 6 meq of the base. Calculate the percentage of magnesium burnt to the nitride.
Answer» Solution :`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%`