AMU Engineering 2020 solved paper Solved Paper

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