Find the limiting reagent and calculate the amount of product formed and reactant left in the following reaction
N2+H2- NH3
if 50g of nitrogen react with 20g of hydrogen
Answers
Answer:
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Explanation:
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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.