10 kg of H2 is burnt in 50 kg of O2 in a closed vessel to produce H20. (a) Which is the
limiting reagent (b) Calculate the amount of one of the reactant which remains
unreacted. (c) Calculate the maximum amount of water that can be produced.
Answers
Explanation:
Limiting reagent → The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.
Given:-
Weight of N
2
=50kg=5×10
4
g
Weight of H
2
=10kg=10
4
g
Molecular weight of N
2
=28g
Molecular weight of H
2
=2g
As we know that,
No. of moles =
Mol. wt.
Weight
Therefore,
No. of moles of N
2
=
28
5×10
4
=1.786×10
3
moles
No. of moles of H
2
=
2
10
4
=5×10
3
moles
Now,
N
2
+3H
2
⟶2NH
3
From the above reaction,
1 mole of N
2
reacts with 3 moles of H
2
to produce 2 mole of ammonia.
Therefore,
1.786×10
3
moles of N
2
will react with 5.36×10
3
moles of H
2
to produce ammonia.
But given amount of H
2
is 5 moles.
Thus H
2
is limiting reagent.
Therefore,
Amount of ammonia produced by 3 moles of H
2
=2 moles
∴ Amount of ammonia produced by 5×10
3
moles of H
2
=
3
2
×5×10
3
=3.33×10
3
moles
Hence 3.33 moles of ammonia gas is formed.
Answered By
toppr
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