$(a)\;27.27\%\qquad(b)\;36\%\qquad(c)\;42\%\qquad(d)\;41.2\%$

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Step 1:

$2Mg+O_2\rightarrow 2MgO$

Before reaction:

$2Mg\Rightarrow a$

$2MgO\Rightarrow 0$

After reaction:

$2Mg\Rightarrow 0$

$2MgO\Rightarrow a$

$3Mg+N_2\rightarrow Mg_3N_2$

Before reaction:

$3Mg\Rightarrow b$

$Mg_3N_2\Rightarrow 0$

After reaction:

$3Mg\Rightarrow 0$

$Mg_3N_2\Rightarrow \large\frac{b}{3}$

Step 2:

Now $(a+\large\frac{b}{3})$ millimole of MgO and $Mg_3N_2$ are present in the mixture.

$MgO+2HCl\rightarrow MgCl_2+H_2O$

$Mg_3N_2+8HCl\rightarrow 3Mgcl_2+2NH_4Cl$

Or the solution contains a millimole of $MgCl_2$ from MgO

b millimole & $\large\frac{2b}{3}$ millimole of $NH_4Cl$ from $Mg_3N_2$

$2a+\large\frac{8b}{3}$ millimole of HCl used .

2a for Mgo

$\large\frac{8b}{3}$ for $Mg_3N_2$

Meq of HCl=60-12=48

$2a+\large\frac{8b}{3}$$=48$

Step 3:

millimole of $NH_4Cl$=millimole of $NH_3$ liberated =millimole of HCl used for absorbing $NH_3$

$\large\frac{2b}{3}$$=4$ or $b=6$

$2a+\large\frac{8\times 6}{3}$$=48$

$a=16$

$\%$ of Mg used for $Mg_3N_2=\large\frac{6}{(6+16)}$$\times 100$

$\Rightarrow 27.27\%$

Hence (a) is the correct answer.

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