$N X$ is produced by the following step of reactions:
1. $M +$ $X_{2}$ $⟶$ $M X_{2}$
2. $M X_{2}$ + $X_{2}$ $⟶$ $M_{3} X_{8}$
3. $M_{3} X_{8}$ + $N_{2} C O_{3}$ $⟶$ $N X +$ $C O_{2}$ + $M_{3} O_{4}$
How much M (metal) is consumed to produce $206 g$ of $N X$ . (Take atomic weight of $M = 56, N = 23, X = 80$ )

42 g

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56 g

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Solution

Correct option: (B)
Given reaction:
$M + X_{2} ⟶ M X_{2}$ .................(1)
$M X_{2} + X_{2} ⟶ M_{3} X_{8}$ .............(2)
$M_{3} X_{8} + N_{2} C O_{3} ⟶ N X + C O_{2} + M_{3} O_{4}$ ............(3)
Multiply the equation (1) by 3
$3 M + X_{2} ⟶ 3 M X_{2}$ ................(4)
Add the equation (1), (2) and (3)
Resultant reaction is: $3 M + 4 X_{2} + N_{2} C O_{3} ⟶ N X + C O_{2} + M_{3} O_{4}$
Given weight of $N X = 206 g$
Molecular mass of $N X = 103 g$

So, Number of moles of $N X$ to be produced = $\frac{206}{103} = 2 m o l e$
From stoichiometry,
$1$ moles of $N X$ is produced from $= 3$ mole of metal
$2$ mole of $N X$ is produced from $= 3 \times 2 = 6$ mole of metal
Hence, mass of metal used to produce $206 g$ of $N X = 6 \times 56 = 336 g$
Hence, the correct option is (B).

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