Question

10 g of N2 treated with 30 g of H2. Calculate limiting reagent in the reaction to form ammonia.​

it's really urgent.

Answer 1

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.