224 lt.of ammonia undergoes catalytical oxidation in presence of Pt to give nitric oxide and water vapour .Calculate thvolume of oxygen required for the reaction.All the volumes measured at room temperature and pressure.

Asked by Sayantany mukhopadhyay | 2nd Aug, 2014, 09:23: AM

Expert Answer:

Reaction:

4NH3(g)   +   5O2(g)   →   4NO(g)   +   6H2O(l)

  4vol.            5vol.            4vol.            6vol.

 

Volume of H2O will be negligible as it is in the liquid form at room temperature.

(Note: Gay Lussac’s Law is not applicable to substances in liquid and solid states).

 

According to Gay Lussac’s Law,

4vols. of NH3 require 5vols. Of oxygen

4 cm3 of NH3 require 5 cm3 Of oxygen

 

Therefore, 224 cm3 of NH3 require oxygen = 5/4 × 224 = 280 cm3

Hence, volume of oxygen required = 280 cm3

 

4vols. of NH3 produce 4vols of NO

4 cm3 of NH3 produce 4 cm3 of NO

Therefore 224 cm3 of NH3 produce NO = 4/4 × 224 = 224 cm3

Hence, volume of NO produced = 224 cm3

Answered by Hanisha Vyas | 4th Aug, 2014, 11:56: AM