Nitrogen forms NO and `NH_3`. Show that equivalent mass is not a fixed quantity and that it varies from reaction to reaction.

Explanation

`N_(2)(28 g )+O_(2) (32 g) rarr 2NOimplies 28 g" nitrogen "= 32 g" oxygen "`.
Equivalent mass of nitrogen in `NO = 7 g ""eq^(-1)`
`N_(2) (28 g) + 3H_(2) (6g) rarr 2NH_(3) implies 28g" nitrogen "= 6g" hydrogen "`
Equivalent mass of nitrogen in `NH_(3) =4.66g eq^(-1)`.