1.

N_(2) + 3H_(2) to 2NH_(3) molecular weight and equivalent weight of NH_(3) and N_(2) are 17.03 g,14 g and Y_(1),Y_(2)respectively. Find the value of (Y_(1)-Y_(2)).

Answer»


Solution :`6e^(-) + N_(2) to 2N^(3-)`
`THEREFORE E_(N_(2)) = 14/6 =Y_(2)`
`therefore E_(NH_(3)) = (17.03)/3 = Y_(1)`
`therefore Y_(1)-Y_(2) = (17.03)/3- X_(2)/6= 5.67 - 2.33 = 3.34`


Discussion

No Comment Found

Related InterviewSolutions