Question

N2 and H2 are mixed in 14:3 mass ratio.After certain time ammonia was found to be 40% by mole. find the mole fraction of N2 at that time in the mixture of N2,H2 and NH3​

Answer 1

The mole fraction of N2 = 3/16, H2 = 9/16 and NH3 = 1/4 respectively.

Explanation:

                                    N2         +       3H2 --->            2NH3

Initial concentration:     x                     3x                       0

At equilibrium:            x[1/40/100]      3[1/40/100]      2 x 40 / 100

                                    x (0.6)               1.8 x                   x (0.8)

Mole fraction of

1) N2:   0.6 / 0.6 x + 1.8 x + 0.8x = 0.6 / 3.2 x = 3 / 16

2) H2:    1.8x / 3.2 x = 9/16

3) NH3:  0.8 x / 3.2 x = 1/4

Thus the mole fraction of N2 = 3/16, H2 = 9/16 and NH3 = 1/4 respectively.

Also learn more

The mole fraction He is 0.4 in gaseous mixture He  and CH4. If both the gaseous are effusing through  the constant area of the orifice of the container, then  what will be % composition by volume of CH4, gas  effusing out initially?

https://brainly.in/question/11773375

Answer 2

Answer:0.15

Explanation:

N2 + 3H2 -› 2NH3

n(N2)=(14m/28 -x )

n(H2)=(3m/2 -3x)

n(NH3)=(2x)

Total number of moles= n(N2)+ n(H2)+n(NH3) = 2m - 2x

According to question

n(NH3)/total no. moles = 40%

Therefore on solving we get

x/m = 2/7

Therefore x= 2m/7

Now mole fraction of N2= (no. of moles of N2)/( total moles)

Mole fraction= (m/2 - x)/(2m-2x)

Putting x=2m/7 we get...

Mole fraction= 3/20 which is equal to 0.15.