For the reaction, `N_(2)(g)+3H_(2)(g)to 2NH_(3)(g)`, if molecular masses of `NH_(3) " and " N_(2) " and " M_(1) " and" M_(2)`, their equivalent masses are `E_(1) " and" E_(2)`, then `(E_(1)-E_(2))` is :

Options

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

Answer

A

Explanation

Equivalent mass of `N_(2), i.e., E_(2)=(M_(2))/(6)`
Equivalent mass of `NH_(3), i.e., E_(1)=(M_(1))/(3)`
Then, `" " E_(1)-E_(2)=(M_(1))/(3)-(M_(2))/(6)=(2M_(1)-M_(2))/(6)`