Question

nitrogen and hydrogen combines in a mass ratio 14:3. calculate the amount of NH3 formed if 42 gram of nitrogen reacts with 12 gram of hydrogen and also identify the limiting reageant.​

Answer 1

Answer:

The molar mass of ammonia is 17 g/mol.

68 grams of ammonia corresponds to 17g/mol68g=4moles

4 moles of ammonia will be obtained from 24×1=2  moles of nitrogen and 24×3=6  moles of hydrogen.

The molar masses of nitrogen and hydrogen are 28 g/mol and 2 g/mol respectively.

2 moles of nitrogen corresponds to 2×28=56  grams.

6 moles of hydrogen corresponds to 6×2=12  grams.

Answer 2

The balanced equation of the

reaction is :-

N₂ + 3H₂ → 2NH₃

Number of moles of N₂ :-

= Given Mass/Molar mass

= 42/28

= 1.5 moles

Number of moles of H₂ :-

= Given Mass/Molar mass

= 12/2

= 6 moles

Hence, we are given more moles of H₂ than N₂ , therefore H₂ is in excess and some of it will remain unreacted, when the reaction is over.

So, N₂ is the limiting reagent and will control the amount of product.

Now, from the reaction :-

∵ 28g N₂ → 34g NH₃

∴ 42g N₂ → 42×34/28 = 51g NH₃

Thus, 51g of NH₃ is formed.