50g of Nitrogen gas and 10g of hydrogen gas are mixed to produce ammonia. Calculate the ammonia formed.Identify the limiting reagent in the production of ammonia in this solition.
Answers
Answer:
Explanation:
N
2
+3H
2
⟶2NH
3
Molecular mass of Nitrogen =28g/mol=0.028kg/mol
Molecular mass of Hydrogen =6g/mol=0.006kg/mol
Molecular mass of Ammonia =17g/mol=0.017kg/mol
Now, according to the balanced chemical equation,
0.028 kg of Nitrogen reacts with 0.006 kg of Hydrogen.
∴ 50 kg of Nitrogen reacts with [
0.028
(0.006×50)
]=10.71kg of Hydrogen.
The amount of Hydrogen (given 10 kg) is less than the amount required (i.e., 10.71 kg) for 50 kg of Nitrogen.
Therefore, Hydrogen is the limiting reagent.
Hence, the formation of Ammonia will depend on the amount of Hydrogen available for reaction.
∵ Amount of ammonia produced by 0.006 kg of hydrogen =2×0.017=0.034kg
∴ Amount of ammonia produced by 10 kg of Hydrogen =
0.006
0.034×10
=56.67kg
Hence, 56.67 kg of ammonia gas will be formed and hydrogen will be limiting reagent.