NX is produced by the following step of reactions

 

M + X₂   ------>  MX ₂

MX₂ + X₂ ------>  M₃X ₈

M₃X₈ + N₂CO₃ ------>   NX + CO₂ + M₃O ₄

How much M (metal) is consumed to produce 206gm of NX. (Take at wt of M=56, N=23, X=80)

 

Dear student!
We have ,

M₃X₈ + N₂CO₃ ---> NX + CO₂ + M₃O₄ 

Here, we can see that, 1 mole of M₃X₈ produces = 1 mole of NX 

Since molecular weight of M₃X₈ = 56 x3 + 80 x 8 = 808  and, NX = 23 + 80 =103 

So, we can say that, 103 gm of NX is produced by = 808 gm of M₃X₈  

So, 206 gm NX will be produced by = 808/103 x 206= 1616gm of M₃X₈  

Now, as 808 gm of M₃X₈ contains = 168 gm M metal , 

so, 1616 gm of M₃X₈ will contain = 168/808 x 1616 = 336 gm of metal.

Hence, 336 gm of metal will be consumed to produce 206 gm of NX compound.