NX is produced by the following step of reactions M + $X_{2}$ $⟶$ $M X_{2}$ $⟶$ $M X_{2}$ + $X_{2}$ $⟶$ $M_{3} X_{8}$

$M_{3} X_{8}$ + $N_{2} C O_{3}$ $⟶$ NX + $C O_{2}$ + $M_{3} O_{4}$ How much M (metal) is consumed to produce 206 gm of NX. (Take at wt. of M = 56, N = 23, X = 80)

42 g

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56 g

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$\frac{14}{3} g$

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$\frac{7}{4} g$

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Solution

The correct option is A
42 g

M+ $X_{2}$ $⟶$ $M X_{2}$

n mole nmole

3 $M X_{2}$ + $X_{2}$ $⟶$ $M_{3} X_{8}$

nmole $\frac{n}{8}$ mole

$M_{3} X_{8}$ + $N_{2} C O_{3}$ $⟶$ 8NX+4 $C O_{2}$ + $M_{3} O_{4}$

$\frac{n}{3}$ $\frac{8 n}{3}$ mole

No. of moles of NX to be produced = $\frac{206}{103}$ =2mole

$\frac{8 n}{3}$ $⟹$ $\frac{3}{4}$

Mass of M = $\frac{3}{4}$ $\times$ 56 =42gm

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Similar questions

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

$N X$ 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 $M = 56, N = 23, X = 80$ )

Q. NX is produced by the following set of reactions.

M + X_{2} \to M X_{2}

3 M X_{2} + X_{2} \to M_{3} X_{8}

M_{3} X_{8} + N a_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}

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

[Take atomic wt. of M = 56, N = 23, X = 80]

NX is produced by the following step of reactions
$M + X_{2} \to M X_{2}$
$3 M X_{2} + X_{2} \to M_{3} X_{8}$
$M_{3} X_{8} + N_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$
How much metal(M) is consumed to produce 206 g of NX? (at wt of M = 56, N = 23, X = 80)