The relation used for formation of ammonia is N2(g)+3H2(g)⇌2NH3(g) Relation by mass =28:6:34 (or) 14:3:17 Given, N2(g)=1.4g⇒H2(g)=1g Thus, N2(g) behave as a limiting reagentand NH3(g) will formed as per amount of N2 used. Therefore, 14g of N2 give NH3=17g ∴1.4g of N2 give NH3=
17×1.4
14
Amount of NH3(g)=1.7g No. of moles of NH3,n=
w
M
n=
1.7
17
=0.1mol Also No. of moles of NH3≡ No. of atoms of N in NH3∴ If no. of moles of NH3=0.1 Then, no. of moles of N-atoms in NH3=0.1mol Hence, No. of moles of NH3=0.1mol No. of moles of N-atoms =0.1mol