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

So at the time of when 0.4 moles of ([tex]NH_{3}[/tex]) is been formed it requires 0.4 moles of ([tex]N_{2}[/tex]) and 3.4 moles of ([tex]H_{2}[/tex])

So, we find the the remaining ([tex]N_{2}[/tex]) will be 0.6 and

([tex]H_{2}[/tex]) will be 0.3 mole present in mixture.

So, the mole fraction of ([tex]N_{2}[/tex]) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

obasola1 obasola1 · Answer 2 · 2020-01-02T01:39:03+01:00

Answer:

0.4615

Explanation:

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

Thus N2 and H2 Mole ratio 1:1.5

H2 is known to be a limiting reagents because it slows down the rate of chemical reaction. when 0.4 mole of NH3 is formed then 0.4 mole N2 and 3×.4 mole of h2 reacts .the remaining n2 is .6 &h2 is 0.3 mole present in mixture.

the remaining N2 =1-0.4=0.6

H2=1.5-1.2=0.3

[tex]N_{2} +H_{2} -------------->NH_{3}[/tex]

balancing the equation

[tex]N_{2} +3H_{2} -------------->2NH_{3}[/tex]

mole fraction of N2=mole fraction/total mole amount

Mole fraction of n2=0.6/(.6+.4+.3)=0.4615