50kg of nitrogen and 10kg of hydrogen are mixed to produce ammonia. Calculate the mass of ammonia form. Identify the limiting reagent in producing ammonia.
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Answer:
Hence, 56.67 kg of ammonia gas will be formed and hydrogen will be limiting reagent.
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11th
Chemistry
Some Basic Concepts of Chemistry
Stoichiometry and Stoichiometric Calculations
50kg N2 and 10kg H2 are mix...
CHEMISTRY
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Asked on December 27, 2019 by
Achugatla Razvi
50kgN
2
and 10kg H
2
are mixed to produce NH
3
. Calculate the ammonia gas formed. Identify the limiting reactant.
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ANSWER
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.