Question

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

Answer 1

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ($N_{2}$)and hydrogen ($H_{2}$)are mixed, the mole ratio becomes 1 : 1.5,

Now we know that ($H_{2}$) is acting as a limiting agent.

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

So, we find the the remaining ($N_{2}$) will be 0.6 and

($H_{2}$) will be 0.3 mole present in mixture.

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

Answer 2

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

$N_{2} +H_{2} -------------->NH_{3}$

balancing the equation

$N_{2} +3H_{2} -------------->2NH_{3}$

mole fraction of N2=mole fraction/total mole amount

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