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N(2) + 3 H(2) rarr 2NH(3) Molecular we...

`N_(2) + 3 H_(2) rarr 2NH_(3)`
Molecular weight of `NH_(3)` and `N_(2)` are `x_(1)` and `x_(2)`, respectively. Their equivalent weights are `y_(1)` and `y_(2)`, respectively. Then `(y_(1) - y_(2))`

A

`((2X_(1)-X_(2))/(6))`

B

`(X_(1)-X_(2))`

C

`(3X_(1)-X_(2))`

D

`(X_(1)-3X_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`6e+N_(2)^(0)rarr2N^(3-)`
`:. E_(N_(2))=(X_(2))/(6)=Y_(2):. E_(NH_(3))=(X_(1))/(3)=Y_(1)`
`:. Y_(1)-Y_(2)=(X_(1))/(3)-(X_(2))/(6)=(6X_(1)-3X_(2))/(18)`
`=(2X_(1)-X_(2))/(6)`
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