NX is produced by the following step of reactionsM+X2→ M X2; 3MX2+X2→M3X8; M3X8+ N2CO3 → NX+CO2+M3O4How much M (metal) is consumed to produce 206 g of NX. (Take at. wt. of M = 56, N = 23, X = 80)

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

Solution:

Balance the given equations to calculate required quantity

$\begin{array}{l} M + X_{2} \to {MX}_{2} \to (1) \times 3 \Rightarrow 3 M + 3 X_{2} \to 3 {MX}_{2} \\ 3 {MX}_{2} + X_{2} \to M_{3} X_{8} \to (2) \\ M_{3} X_{8} + N_{2} {CO}_{3} \to NX + {CO}_{2} + M_{3} O_{4} \to (3) \\ By balancing eq . (3) \Rightarrow M_{3} X_{8} + 4 N_{2} {CO}_{3} \to 8 NX + 4 {CO}_{2} + M_{3} O_{4} \\ From the above equations; \\ 3 moles of' M' \to 8 moles of NX \\ 3 \times 56 g of' M' \to 8 (23 + 80) g of NX \\ ? \to 206 g of NX \\ mass of' M' consumed = \frac{206}{8 (103)} \times 3 \times 56 = 42 g \end{array}$