`N_(2) + 3 H_(2) rarr 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))`A. `((2X_(1)-X_(2))/(6))`B. `(X_(1)-X_(2))`C. `(3X_(1)-X_(2))`D. `(X_(1)-3X_(2))`

`N_(2) + 3 H_(2) rarr 2NH_(3)` Molecular weight of `NH_(3)` and `N_(2)

1.	`N_(2) + 3 H_(2) rarr 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))`A. `((2X_(1)-X_(2))/(6))`B. `(X_(1)-X_(2))`C. `(3X_(1)-X_(2))`D. `(X_(1)-3X_(2))`
Answer» Correct Answer - A `6e+N_(2)^(0)rarr2N^(3-)` `:. E_(N_(2))=(X_(2))/(6)=Y_(2):. E_(NH_(3))=(X_(1))/(3)=Y_(1)` `:. Y_(1)-Y_(2)=(X_(1))/(3)-(X_(2))/(6)=(6X_(1)-3X_(2))/(18)` `=(2X_(1)-X_(2))/(6)`

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`N_(2) + 3 H_(2) rarr 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))`A. `((2X_(1)-X_(2))/(6))`B. `(X_(1)-X_(2))`C. `(3X_(1)-X_(2))`D. `(X_(1)-3X_(2))`

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