`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))`

Options

A. `((2x_(1) - x_(2))/(6))`
B. `(x_(1) - x_(2))`
C. `(3x_(1) - x_(2))`
D. `(x_(1) - 3x_(2))`

Answer

A

Explanation

`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))`