Question

10g hidrogen and 50g nitrogen are allowed to react to give ammonia. identify the limiting reagent and calculate the mass of ammonia formed​

Answer 1

Answer:

56.67 kg

Explanation:

N2+3H2⟶2NH3

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.

Answer 2

Answer:

1. ycyovy ry hu to pitch gyx vivuut

Explanation:

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