# 50kg of nitrogen and 10 kg of hydrogen are mixed to produce ammonia . Calculate the ammonia formed and identify the limiting reagent in the production of ammonia in this situation.

By Shivank Goel|Updated : August 12th, 2022

We know that

N2 + 3H2 --> 2NH3

Moles of nitrogen = 50/28 = 25/14 = 1.78

Moles of hydrogen = 10/2 = 5

1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of NH3

1.78 mole of nitrogen reacts with 1.78 x 3 = 5.34 moles of Hydrogen

So hydrogen is the limiting reagent.

1 mole of hydrogen will give ⅔ moles of NH3

5 moles of hydrogen will give 10/3 moles of NH3

Moles of ammonia formed = 10/3 = 3.33 mol

Therefore, the ammonia formed is 3.33 mol and the limiting reagent in the production of ammonia in this situation is hydrogen.

Summary:

## 50kg of nitrogen and 10 kg of hydrogen are mixed to produce ammonia. Calculate the ammonia formed and identify the limiting reagent in the production of ammonia in this situation.

50kg of nitrogen and 10 kg of hydrogen are mixed to produce ammonia. The ammonia formed is 3.33 mol and the limiting reagent in the production of ammonia in this situation is hydrogen.

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