Question

If 5g H2, is mixed with 14g of nitrogen gas for the following reaction N2 + 3H2 → 2NH3 At the end , mass of H2, left unreacted is?​

Answer 1

Answer:

2g

Explanation:

N2 + 3H2 → 2NH3

28g      6g        34g

28g N2 react with H2 = 6g

14g N2 react with H2= (6x14/28) = 3g H2

total H2 = 5g

react H2 = 3g

left unreacted H2 = 5g - 3g = 2g

Answer 2

Answer: Thus, 14gN2 produces 17gNH3 and the unreacted H2 amount is 2g

Explanation:

The balanced equation of hydrogen reacting with nitrogen gas to give ammonia is N2+3H2→2NH3. From this equation we notice that 3 mol of H2 is required with 1 mol of nitrogen gas to produce 2 moles of NH3. Therefore, 3×2=6gH2 ,

1×28=28gN2 and

2×17=34gNH3

It is given to us that

N2=14g

H2=5g

So, we know that,

28g N2→6g H2

⇒1g N2→628g H2

So, 14g N2→628×14g H2

⇒14g N2→3g H2

We are given 5g of H2 in the question thus, we now know that H2 is in excess and is not the limiting reagent. But we also have 14gN2 that is not sufficient for complete reaction with 5gH2 and thus, N2 is the limiting reagent.

Thus, the given amount of N2 gets used up completely for the reaction but some H2 is left unreacted. So, the unreacted H2 left is 5−3=2g .

Now for calculating the amount of ammonia produced by the reaction of the given amount of hydrogen and nitrogen gas will depend on the amount N2 because it is the limiting reagent. From the question we know that

28g N2→34g NH2

⇒1g N2→3428g NH3

⇒14g N2→3428×14g NH3

Thus, 14gN2 produces 17gNH3 and the unreacted H2 amount is 2g

Note:

The production of ammonia NH3 is known as Haber-Bosch process. In this process natural gas or LPG is passed into an ammonia producing plant where gaseous hydrogen is collected and this hydrogen is reacted with nitrogen gas to produce ammonia as a product of their reaction.

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