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 t

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victus00

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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 t

ime in the mixture of N2, H2, and NH3 is

Chemistry

2 answers:

lorasvet [2.7K] · Answer 1 · 2026-02-21T12:11:23+03:00

Answer: The mole fraction of nitrogen is calculated to be 0.4615.

Explanation: In the mixture of nitrogen ( $N_{2}$ ) and hydrogen ( $H_{2}$ ), the mole ratio established is 1: 1.5.

At this juncture, we identify that ( $H_{2}$ ) serves as the limiting reagent.

When 0.4 moles of ( $NH_{3}$ ) are produced, it will require 0.4 moles of ( $N_{2}$ ) along with 3.4 moles of ( $H_{2}$ ).

The total amount of remaining ( $N_{2}$ ) ends up being 0.6, while the amount of ( $H_{2}$ ) left in the mixture is 0.3 moles.

The mole fraction of ( $N_{2}$ ) is computed as 0.6 divided by the sum of 0.6, 0.4, and 0.3, resulting in 0.4615.

eduard [2.7K] · Answer 2 · 2026-02-21T12:11:27+03:00

Answer:

0.4615

Explanation:

N2 and H2 are combined in a 14:3 mass ratio. After some duration, ammonia was determined to be 40% by mole. The mole fraction of N2 during that time in the combination of N2, H2, and NH3 is

The mole ratio of N2 to H2 is 1:1.5.

H2 is identified as a limiting reagent due to its effect on slowing the chemical reaction rate. Upon forming 0.4 mole of NH3, 0.4 mole of N2 and 3×0.4 moles of H2 will react. Thus, the remaining amounts are 0.6 moles of N2 and 0.3 moles of H2 in the mixture.

The leftover N2 is calculated as 1-0.4=0.6.

The remaining amount of H2 is 1.5-1.2=0.3.

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