50.0 gm of N₂ (g) and 10.0 gm of H₂ (g) are mixed to produce NH₃ (g). Calculate the NH₃ (g) formed. Identify the limiting reagent in the production of NH₃ :[Given : Gram atomic mass of N= 14 gm] *
H₂ is Limiting Reagent and approx. 28 gm of Ammonia is formed
H₂ is Limiting Reagent and approx. 56 gm of Ammonia is formed
N₂ is Limiting Reagent and approx. 28 gm of Ammonia is formed
N₂ is Limiting Reagent and approx. 56 gm of Ammonia is formed
Answers
Answer:
Explanation:
As we know that,
no. of moles=
mol. wt.
Wt.
Weight of N
2
=50kg
Molecular weight of N
2
=28g
No. of moles of N
2
=
28
50×10
3
=17.86×10
2
moles
Weight of H
2
=10kg
Molecular weight of N
2
=2g
No. of moles of N
2
=
2
10×10
3
=5×10
3
moles
N
2
+3H
2
⟶2NH
3
From the above reaction,
1 mole of N
2
react with 3 moles of H
2
.
No. of moles of H
2
required to react with 17.86×10
2
moles of N
2
=3×17.86×10
2
=5.36×10
3
moles
But only 5×10
3
moles of H
2
are available.
Thus H
2
is the limiting reagent here.
Now, again from the above reaction,
Amount of ammonia formed when 3 moles of H
2
react =2 moles
Therefore,
Amount of ammonia formed when 5×10
3
moles of H
2
react =
3
2
×(5×10
3
)=3.33×10
3
moles
Molecular weight of ammonia =17g
Weight of ammonia in 3.33×10
3
moles =17×(3.33×10
3
)=56.61kg.
Hence the amount of NH
3
formed is 56.61kg.