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 BYJU'S Exam Prep
Updated on: September 13th, 2023
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.
Related Questions:-
