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

42 g m

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336 g m

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\frac{14}{3} g m

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\frac{7}{4} g m

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Solution

The correct option is C $336 g m$
$M + X_{2} \to M X_{2}$
$3 M X_{2} + X_{2} \to M_{3} X_{8}$
$M_{3} X_{8} + N_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$

Multiplying Equation 1 by Equation 3, and adding all the equations we get
$3 M + 3 X_{2} \to 3 M X_{2}$
$3 M X_{2} + X_{2} \to M_{3} X_{8}$
$M_{3} X_{8} + N_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$
Resultant equation:- $3 M + 4 N_{2} + N_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$
Mass of $N X = 206 g$
Molar mass of $N X = 23 + 80 = 103 g / m o l$
Moles of $N X = 206 g \times \frac{1 m o l}{103 g} = 2 m o l N X$
Moles of metal $M = 2 m o l N X \times \frac{3 m o l e M}{1 m o l e N X} = 6 m o l M$
Mass of metal $M = 6 m o l M \times \frac{56 g M}{1 m o l e M} = 336 g M$
Hence, 336g metal is consumed to produce 206g of $N X$ .

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Similar questions

$N X$ 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 + $X_{2}$ $⟶$ $M X_{2}$ $⟶$ $M X_{2}$ + $X_{2}$ $⟶$ $M_{3} X_{8}$ $M_{3} X_{8}$ + $N_{2} C O_{3}$ $⟶$ 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)

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.

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

M

(metal) is consumed to produce

206 g

N X

?
(Take molar mass of

M = 56, N = 23, X = 80 g/mol

)

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NX is produced by the following step of reactions:M+X2⟶MX23MX2+X2⟶M3X8M3X8+N2CO3⟶NX+CO2+M3O4How much M (metal) is consumed to produce 206gm of NX?(Take atomic weight of M=56,N=23,X=80)

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