Consider the following reaction , `N_(2) + 3H_(2) to 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))` is

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`
eq. mass of `N_(2)=(x_2)/(6)=y_(2)" "` eq. mass of `NH_(3)=(2x_1)/(6)=y_(1)`
1 eq. Of `N_2` combines with 1 eq. Of `H_2` to form 1 eq. Of `NH_3`
`therefore (y_(1)-y_(2))=((2x_1)/(6)-(x_2)/(6))=((2x_(1)-x_(2))/(6))`.