A sample of Mg 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 was back titrated with NaOH .12 Meq . of NaOH were requi

Question

A sample of Mg 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 was 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 Mg burnt to the nitride .

TestHub · Accepted Answer

Total HCl taken = 60 meq, NaOH used in first back titration = 12 meq ⇒ HCl consumed = 48 meq.
From NH₃ estimation: acid taken = 10 Meq, base used = 6 meq ⇒ NH₃ neutralized acid = 4 meq.
NH₃ comes only from Magnesium nitride, so 4 Meq corresponds to N³⁻ from Mg₃N₂.
1 mole Mg₃N₂ gives 2 NH₃ ⇒ equivalents relation gives Meq of Mg₃N₂ = 12 Meq (after stoichiometry).
Out of total 48 Meq Mg reacted, 12 Meq formed nitride.
% Mg → nitride = (12 / 44) × 100 = 27.27%

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