2.8 kg of N2 gas and 2kg of H2 gas are mixed to produce NH3 gas. a) Calculate the amount of NH3 formed. b) Identify the limiting reagent in this reaction. c) Calculate the amount of unreacted reactant
Answers
Answer:
your answer is 3.33
Explanation:
Solution:-
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.
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