Question

Pure nitrogen is obtained by decomposition of nitride ${{\text{M}}_{\text{x}}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}}$ where, ${\text{x + y = }}$ ?

Answer 1

Hint In the above question ${{\text{M}}_{\text{x}}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}}$ compound is given where ${\text{x}}$ & ${\text{y}}$ are the molecularity of the element M and ${{\text{N}}_{\text{3}}}$ molecule respectively present in the given compound. And for knowing the value of ${\text{x}}$ & ${\text{y}}$ we have to write the balanced chemical equation of the decomposition of nitride.

Complete step by step solution:
Decomposition reaction for the formation of pure nitrogen in the form of gas is written as below:
${\text{Na}}{{\text{N}}_{{\text{3}}\left( {\text{s}} \right)}} \to {\text{N}}{{\text{a}}_{\left( {\text{s}} \right)}}{\text{ + }}{{\text{N}}_{{\text{2}}\left( {\text{g}} \right)}}$
In the above given decomposition chemical equation, sodium azide ( ${\text{Na}}{{\text{N}}_{\text{3}}}$ ) decompose to form nitrogen gas ( ${{\text{N}}_{\text{2}}}$ ) and produces sodium (${\text{Na}}$ ) in the solid form. Above equation is not present in the balanced form because molecularity of each atom is not equal on the left hand side and on the right hand side of the chemical equation, so balanced form of the above reaction is shown as follow:
${\text{2Na}}{{\text{N}}_{{\text{3}}\left( {\text{s}} \right)}} \to 2{\text{N}}{{\text{a}}_{\left( {\text{s}} \right)}}{\text{ + 3}}{{\text{N}}_{{\text{2}}\left( {\text{g}} \right)}}$
-Here sodium azide ( ${\text{Na}}{{\text{N}}_{\text{3}}}$ ) decomposes to form pure nitrogen gas it means the compound which satisfies the molecular formula of ${{\text{M}}_{\text{x}}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}}$ is sodium nitride i.e.
${{\text{M}}_{\text{x}}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}} = {\text{Na}}{{\text{N}}_{\text{3}}}$
Or ${{\text{M}}_{\text{x}}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}} = {\left( {{\text{Na}}} \right)_x}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}}$
Or ${\left( {{\text{Na}}} \right)_x}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{y}}} = {\left( {{\text{Na}}} \right)_1}{\left( {{{\text{N}}_{\text{3}}}} \right)_1}$
From above data it is clear that the value of ${\text{x}}$ and ${\text{y}}$ is one.
-And here in this question we have to calculate the value of ${\text{x + y}}$ which is equals to two, and obtain by putting the values of ${\text{x}}$ and ${\text{y}}$ equals to one.
i.e. ${\text{x + y = 1 + 1 = 2}}$ .

Note: Here some of you may do wrong calculation if you think that value of ${\text{y}}$ is three by only seeing three in the nitrogen molecule ( ${{\text{N}}_{\text{3}}}$ ) of sodium azide in the above decomposition reaction.