amonia mole one n2 and h3 now amonia symbol and naitrogen jojota?
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Given reaction for the formation of ammonia-
N
2
+3H
2
⟶2NH
3
From the above reacton,
1 mole of N
2
reacts with 3 moles of H
2
.
Therefore,
6 moles of N
2
will react with 18 moles of H
2
.
Given amount of H
2
=6 moles
Therefore, H
2
is the limiting reagent.
Therefore,
Again from the above reaction.
No. of moles of ammonia formed when 3 moles of H
2
react with excess of N
2
=2
∴ No. of moles of ammonia formed when 6 moles of H
2
react with excess of N
2
=
3
2
×6=4 moles
Hence 4 moles of ammonia are formed when 6 moles of N
2
react with 6 moles of H
2
.
Answer ByavatarToppr
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