50.0 kg of N (g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the moles of
NH3(g) formed. Identify the limiting reagent in the production of NH3 in this situation.
Hint : N2(g) + 3H2(g) –
2NH3(g)
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.
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