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

0.4615

Step-by-step 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

Answer 2

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