NX is produced by the following step of reactions M - X 2- M X 23 M X 2- X 2- M 3 X 8M 3 X 8- N 2 CO 3- NX - CO 2- M 3 O 4How much metalM is consumed to produce 206 g of NX- at wt of M -56- N -23- X -80 A- 42 gB- 56 gC- 14-3 gD- 7-4 g

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Byju's Answer

Standard XII

Chemistry

Potassium Dichromate

NX is produce...

Question

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)

$42 g$

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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
$42 g$

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

This reaction is not balanced. On balancing, you get:

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

206 g is 2 mol of NX.

3 mol M gives us 8 moles NX, how much will give us 2 mol?

$\frac{3}{4}$ mol, = $\frac{3}{4} \times 56 = 42 g$

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

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

)

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

)

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)

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NX is produced by the following step of reactions M+X2→MX2 3MX2+X2→M3X8 M3X8+N2CO3→NX+CO2+M3O4 How much metal(M) is consumed to produce 206 g of NX? (at wt of M = 56, N = 23, X = 80)

The correct option is A 42g M3X8+N2CO3→NX+CO2+M3O4 This reaction is not balanced. On balancing, you get: M3X8+4N2CO3→8NX+4CO2+M3O4 206 g is 2 mol of NX. 3 mol M gives us 8 moles NX, how much will give us 2 mol? 34 mol, = 34×56=42 g

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)