NX is produced by the following step of reactions M+X2→MX2 3MX2+X2→M3X8 M3X8+N2CO3→NX+CO2+M3O4How much M (metal in grams) is consumed to produce 206 g 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
   $M + X_{2} \to M X_{2}$
   $3 M X_{2} + X_{2} \to M_{3} X_{8}$
   $M_{3} X_{8} + N_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$
How much M (metal in grams) is consumed to produce 206 g of NX? (Take at. Wt. of M = 56,
n = 23, X = 80)

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answer is 42.

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Detailed Solution

$M + X_{2} \to M X_{2} M X_{2} + X_{2} \to M_{3} X_{8} M_{3} X_{8} + N a_{2} C O_{3} \to N X + C O_{2} + M_{3} O_{4}$
Mole of NX = $\frac{206}{103} = 2$
POAC for X atom
No. of atom in $M_{3} X_{8}$ = No. of X atom in NX
8[No. of mole of $M_{3} X_{8}$ ] = 1[No. of mole of NX]
No. of mole of $M_{3} X_{8}$ = $[\frac{2}{8}] = \frac{1}{4}$ mole
Now, POAC for M atom
3[No. of mole of $M_{3} X_{8}$ ] = 1 x [N0. Of mole of M]
$∴$ 3 x $\frac{1}{4} =$ No, of mole of M
Weight of M atom = $\frac{3}{4}$ x 56 = 42 gram