Question

Problems 17 4.40. Ammonia is burned to form nitric oxide in the following reaction ANH + 50 + 4NO + 640 (a) Calculate the ratio (Ib-mole O, react/lb mole NO formed). (b) Ifainmonia is led to a continuous reactor at a rate of 1000 kmol NH,/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O,? (c) If 500 kg of ammonia and 100,0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction (mol) and mass of NO produced (kg) if the reaction proceeds to completion d u resaared sulfur very

Answer 1

Ans 4.40

The balanced reaction

4NH3 + 5O2 = 4NO + 6H2O

Part a

From the stoichiometry of the reaction

lbmol O2 react / lbmol NO formed = 5/4 = 1.25

Part b

Moles of NH3 fed = 100 kmol/h

From the stoichiometry of the reaction

Moles of O2 required

= 5 kmol O2 x 100 kmol NH3/h / 4 kmol NH3

= 125 kmol/h

O2 is in 40% excess

O2 feed rate = 125*(100 + 40)% = 175 kmol/h

Part c

Mass of NH3 = 50 kg

Moles of NH3 = mass/molecular weight

= (50 kg) / (17 kg/kmol)

= 2.941 kmol

Mass of O2 = 100 kg

Moles of O2 = mass/molecular weight

= (100 kg) / (32 kg/kmol)

= 3.125 kmol

Moles before reaction = 2.941 + 3.125 = 6.066 kmol

From the stoichiometry of the reaction

4 kmol NH3 reacts with = 5 kmol O2

2.941 kmol NH3 reacts with = 5*2.941/4 = 3.676 kmol O2

But we have less 3.125 kmol O2 than required 3.676 kmol.

Limiting reactant = O2

Excess reactant = NH3

Moles of NH3 reacted

= 4 kmol NH3 x 3.125 kmol O2 / 5 kmol O2

= 2.50 kmol

Excess moles of NH3 = 2.941 - 2.500 = 0.441 kmol

Mass of NH3 in excess = moles in excess x molecular weight

= 0.441 kmol x 17 kg/kmol

= 7.497 kg

% excess = 7.497*100/50 = 15%

Moles of NO produced

= 4 kmol NO x 3.125 kmol O2 / 5 kmol O2

= 2.50 kmol

Mass of NO produced = moles x molecular weight

= 2.50 kmol x 30 kg/kmol

= 75 kg

Moles after reaction = moles of NO + moles of NH3 + moles of H2O + moles of O2

= 2.50 + 0.441 + (6*3.125/5) + 0

= 6.69 kmol

Extent of reaction = (moles after reaction - moles before reaction) / (sum of stoichiometric coefficients)

= (6.69 - 6.066) / (10 - 9)

= 0.624 mol