67.2 litres of H2 combines with 44.8 litres of N2 to form NH3 :
N2(g) + 3H2(g) ⟶ 2NH3(g).
Calculate the vol. of NH3 produced. What is the substance, if any, that remains in the resultant mixture ?

[By Lussac’s law]

N₂     +    3H₂    →    2NH₃

1 vol   :   3 vol   →   2 vol

To calculate the volume of ammonia gas formed.

H₂    :     NH₃

3       :      2

67.2  :     x

⇒ x = 2\over 3 × 67.2 = 44.8 lit

Hence, volume of NH₃ formed is 44.8 lit.

We know,

H₂    :     N₂

3       :     1

67.2  :    x

⇒ x = 1\over 3 × 67.2 = 22.4 lit

∴ nitrogen left = 44.8 – 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.