Question

$NX$ is produced by the following step of reactions:

$M+X_2⟶M{X}_2$

$3M{X}_2+X_2⟶M_3{X}_8$

$M_3{X}_8+N_2{CO}_3⟶NX+{CO}_2+M_3{O}_4$

How much $M$ (metal) is consumed to produce $206gm$ of $NX$?

(Take atomic weight of $M=56,N=23,X=80$)

• A. $42gm$
• B. $\frac{14}{3}gm$
• C. $336gm$
• D. $\frac{7}{4}gm$

Answer 1

Answer: (C)

$M+X_2\to M{X}_2$

$3M{X}_2+X_2\to M_3{X}_8$

$M_3{X}_8+N_{2}C{O}_3\to NX+C{O}_2+M_3{O}_4$

Multiplying Equation 1 by Equation 3, and adding all the equations we get

$3M+3X_2\to 3M{X}_2$

$3M{X}_2+X_2\to M_3{X}_8$

$M_3{X}_8+N_{2}C{O}_3\to NX+C{O}_2+M_3{O}_4$

Resultant equation:- $3M+4N_2+N_{2}C{O}_3\to NX+C{O}_2+M_3{O}_4$

Mass of $NX=206g$

Molar mass of $NX=23+80=103g/mol$

Moles of $NX=206g\times \frac{1mol}{103g}=2molNX$

Moles of metal $M=2molNX\times \frac{3moleM}{1moleNX}=6molM$

Mass of metal $M=6molM\times \frac{56gM}{1moleM}=336gM$

Hence, 336g metal is consumed to produce 206g of $NX$.