Question

$NX$ is produced by the following step of reactions.
$M+X_2\to M{X}_2$
$3M{X}_2+X_2\to M_3{X}_8$
$M_3{X}_8+N_{2}C{O}_3\to NX+C{O}_2+M_3{O}_4$
How much $M$ (metal) is consumed to produce $206\text{g}$ of $NX$?
(Take molar mass of $M=56,N=23,X=80\text{g/mol}$)

• A. $42\text{g}$
• B. $56\text{g}$
• C. $\frac{14}{3}\text{g}$
• D. $\frac{7}{4}\text{g}$

Answer 1

The correct option is A $42\text{g}$

Number of moles of $NX=\frac{\text{given weight}}{\text{molar mass}}=\frac{206}{103}=2\text{mol}$

$M+X_2\to M{X}_2..\left(i\right)$
$3M{X}_2+X_2\to M_3{X}_8..\left(ii\right)$
$M_3{X}_8+4N_{2}C{O}_3\to 8NX+4C{O}_2+M_3{O}_4..\left(iii\right)$

In reaction (iii), 8 mol of $NX$ are produced by 1 mol of $M_3{X}_8$.
2 mol will be produced by = $\frac{1}{8}\times 2$ = 0.25 mol.

In reaction (ii), 1 mol of $M_3{X}_8$ is produced by 3 mol of $M{X}_2$.
0.25 will be produced by = $\frac{3}{1}\times 0.25$ = 0.75 mol.

In reaction (i), 1 mol of $M{X}_2$ is produced by 1 mol of $M$.
0.75 mol will be produced by 0.75 mol of $M$.
Amount of $M=0.75\times 56=42\text{g}$