Question

15. The first step in the production of nitric acid is the oxidation of ammonia over a platinum catalyst to nitric oxide: 4NH₃ + 5O₂ ?4NO + 6H₂O

Under certain reaction conditions we get 90% conversion of NH₃ from a feed of 40 mol/hr NH₃ and 60 mol/hr O₂

Find the output rate of all species from the reactor

Answer 1

4NH3 + 5O2 > 4NO + 6H2O

X = 0.90

FEED = 40 mol/h of NH3 + 60 mol/h O2

Output:

Lets do a mass balance using reaction

Find limiting reactant

40 mol of NH3 will need 50 lbmol of O2 which we have (actually we got 60) so NH3 is limiting and O2 is in excess

base your calculaitons on NH3

40 mol NH3 react 90% = 36 mol of NH3 reacted

Mole Balance for NH3

initial - change = final

40 mol - 36 reacted = 4 mol out NH3

Mole Balance for O2

initial - change = final

60 mol - ? reacted = X mol O2

reacted? 4:5 if we reacted 36 mol of ammonia we need 5/4 of that = 45 mol of O2 used

60 mol - 45 mol reacted = 15 mol of O2 out

Mole Balance for NO

initial + change = final

0 mol + ? produced = outlet of NO

4:4 stoichiometric ratio, since we reacted 36 mol then 36 mol are produced

36 mol of NO outet

Mole Balance for H2O

initial + change = final

0+? produced = mol H2O

by stoichiometry 4:6, for 36 mol we need 6/4 mol of that = 54 mol of H2O is produced

0 + 52 = mol out of H2O

52 mol of H2O in the oultet

In the outlet:

36 mol NH3

15 mol of O2

36 mol of NO

56 mol of H2O

total mol out= 36+15+36+56 = 143 mol