Question

how much volume of NH3 will form if 20 L of H2 is reacting with excess N2 at constant pressure and temperature??​

Answer 1

Answer:

The volume of NH₃ formed is 13.2L

Explanation:

The balanced equation is,

              3H₂ + N₂ → 2NH₃

3 moles of H₂ reacts with 1 mole of N₂ to produce 2 moles of NH₃.

Since N₂ is excess, the reaction depends on the amount of H₂. Thus, H₂ is the limiting reagent of the reaction.

Given,

     volume of H₂ = 20L

then,

       moles of H₂ = 20L× (1mol/22.4L)

⇒                        = 0.89 moles.

          3 moles of H₂ ≡ 2 moles of NH₃.

       0.89 moles f H₂ ≡ 0.89×(2/3)moles of NH₃.

⇒         moles of NH₃ = 0.59mol.

     volume of  NH₃ =  0.59mol × (22.4L/1mol)

⇒                             = 13.2L

Hence, the required answer is 13.2L.