In the above reaction,what is the equivalent weight of NH3 and H2 and N2?
Asked by pb_ckt | 19th May, 2019, 11:59: PM
In formation of ammonia-
Oxidation state of N in N2 =0
Oxidation state of N in NH3 =-3
change in oxidation number of N in N2 in formation of NH3 =3
For 2 N atom = 6
So equivalent weight of N2 =28/6=14/3 g(molecular weight of N2 =28 g)
Equivalent weight of NH3= 17/3 g (molecular weight of NH3 =17 g)
Oxidation number change in H is =1 ( from 0 to +1)
so for two H atom in H2 =2
It means, equivalent weight of H2 = 2/1=2 g
Answered by Ravi | 20th May, 2019, 11:30: AM
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