Question

Ammonia is burned to form nitric oxide in the following reaction: NH3 + O2  NO + H2O

(i) Determine if the above stoichiometric equation is balanced and calculate the ratio of kmol NH3 react/kmol NO formed

(ii) If 25% excess O2 is fed to a continuous reactor at a rate of 250 kmol O2/h, calculate the rate of ammonia fed to a reactor in kmol/min.

(iii) If the above reaction is carried out in a batch reaction with 90 kg of ammonia and 160 kg of oxygen as feed, determine the limiting reactant under this condition and the percentage excess of the other reactant

(iv) Under the condition stated in (iii), calculate the extent of reaction (kmol) and the amount of H2O formed (kg) if the reaction proceeds to completion.

Answer 1

(i) NH3 + O2 --> NO + H2O

the balanced equation will be:

4NH3 + 5O2 --> 4NO + 6H2O

So as per stoichiometry, the Kmol NH3 react / Kmol NO formed = 1:1

(ii) If 25% excess O2 is fed to the reactor at the rate of 250 Kmol O2 / hour

250 = x + (25 /100 )x

250 = x + 0.25x

250 = 1.25x

x = 200

The 25% is used means the actual moles required are = 200 Kmoles / h

The ammonia and oxygen ratio = 4:5

so if 200K moles / h of oxygen are required then the moles of NH3 = 200 X 4 / 5 = 160 Kmoles / h

rate of ammonia fed to a reactor in kmol/min = 160 Kmol /min

(iii) The mass of ammonia = 90 Kg

So moles = 90 / 17 Kmol = 5.3 Kmol

The mass of Oxygen = 160 Kg

So moles of oxygen = 160 / 32 = 5 Kmol

So the excess reagent = ammonia

actual amount of ammonia required = 4

The limiting reagent = oxygen

The excess of ammonia = 5.3 - 4 = 1.3 Kmoles

The % excess = moles excess / moles required = 1.3 X 100/ 4 = 32.5 %

(iv) The reaction will go to 4 Kmole of NH3 reaction with 5 kmol of oxygen , producing 4 kmol of NO and 6 Kmole of H2O

The mass of H2O formed = Moles X molecular weight = 6 Kmoles X 18 = 108 Kg