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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)` and `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
`N_(2) -= 2NH_(3) -= 3H_(2) -= 6H`
`Ew "of" N_(2) = (x_(2))/(6) = y_(2)`
`Ew "of" NH_(3) = (2x_(1))/(6) = y_(1)`
`:.y_(1) = y_(2) = ((2x_(1))/(6) - (x_(2))/(6)) = ((2x_(1) - x_(2))/(6))`
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