Problem 1.5
50.0 kg of N, (g) and 10.0 kg of H, (g) are
mixed to produce NH, (g). Calculate the
NH, (g) formed. Identify the limiting
reagent in the production of NH3 in this
situation.
Answers
Answer:
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