Question

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

Answer 1

The given series of chemical reactions are,

Equation-1: $M + X_{2}-->MX_{2}$

Equation - 2: $3 MX_{2}+X_{2}-->M_{3}X_{8}$

Equation-3: $M_{3}X_{8}+N_{2}CO_{3} -->NX + CO_{2}+M_{3}O_{4}$

Multiplying the equation-1 by 3 and adding all the equations, we get

$3M + 3X_{2}-->3MX_{2}$

$3 MX_{2}+X_{2}-->M_{3}X_{8}$

$M_{3}X_{8}+N_{2}CO_{3} -->NX + CO_{2}+M_{3}O_{4}$

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$3 M + 4 X_{2}+N_{2}CO_{3} -->NX + CO_{2}+M_{3}O_{4}$

Mass of NX = 206 g

Molar mass of NX = 23 + 80 = 103 g/mol

Moles of NX = $206 g * \frac{1 mol}{103 g} = 2 mol NX$

Moles of metal M = $2 mol NX * \frac{3 mol M}{1 mol NX} = 6 mol M$

Mass of metal M = $6 mol M * \frac{56 g M}{1 mol M} = 336 g M$

Therefore, 336 g metal is consumed to produce 206 g NX.

Answer 2

Solution: Given, 1 mole of M3X8 gives 1 mole of NX.

Molecular weight of M3X8 = 56 x 3 +80 x 8= 808 and NX = 23+80 = 103.

Therefore, in order to produce 206 gm of NX, 808/103 x 206 = 1616 gm of M3X8 will be required.

The metal M will be composed as 168/808 x 1616 = 336gm.