Question

Calculate the following quantities for the reaction in which ammonia is formed from its atomic components:

a) The stoichiometric coefficients of the reactants and products.

b)The stoichiometric ratio of N₂ to H₂.

c)(moles NH₃ produced)/(mole N₂ reacted)

d)(lb_m H₂ reacted)/(lbmole NH₃ produced)

e)The Kmole of N₂ and H₂ that must react to form 150 Kmoles of NH₃.

f)The lb_m of NH₃ produced and lb_m of H₂ reacted if 20 lb_m of N₂ react completely.

Answer 1

The reaction is N2+3H2----> 2NH3

The stoichiometric coefficients are N2= 1 H2= 3 and NH3=3

Moles of Ammonia formed/ mole of N2 reacted= 2/3

c) Basis : 2 lbmoles of NH3 produced

2 lb moles of ammonia requires 3 lb moles of H2

Mole = Mass/ Molecular weight

Mass of water= 3*18= 54 lb of water

lb of H2 reacted/ lbmoels of ammonia produced= 54 lb/2 lb moles of NH3= 27 lb hydrogen/ lbmole of NH3

e) for producing 150 kmoles of NH3

N2 required= 150 k moles of N2 and H2= 3*150= 450 k moles

f) Moles of H2= 1/ 2= 0.5 lbmoles Moles of N2= mass/ molecular weight = 20/28= 0.7142

1 lb mole of Nitrogen requires 3 lb moles of hydorgen as per the reaction

Stoichimetric requirement of N2 : H2= 1 :3 = 1/3 : 3= 0.333 :1

The reactants are suppiled at 0.7142; 0.5 = 1: 0.7 =1/0.7 :1 = 1.428 :1

since nitrogen required = 0.33 moles where as supplied is 1.428 for reacting with 1 mole of hydorgen, hyrogen is the limting reactant

As per the reaction, 3 lb moles of hydrogen requires 1 lb mole of Nitrogen

0.5 lb moles of hydrogen requires 0.5/3 lb mole of nitrogen and produces 2*0.5 =1 lb mole of NH3

0.5 lb moles require 0.1667 lb moels of nitrogen and produces 0.5*17 lb of nitroen

lb of nitrogen produced = 0.5*17= 8.5 lbm

0.5 lb moles of hydorgne produces 0.5*2/3=