In the Deacon process for the manufacture of chlorine, HCl and O2 react to form
ID: 950973 • Letter: I
Question
In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole % O2, 79% N2) is fed to provide 35.0% excess oxygen and the fractional conversion of HCl is 75.0%.
a. Calculate the mole fractions of the product stream components using atomic species balances in your calculation.
b. Again calculate the mole fractions of the product stream components using the extent of reaction and stoichiometric coefficients in the calculation for the conditions of 40.0% excess oxygen and a fractional conversion of HCl of 80.0%
Explanation / Answer
The reaction is 4HCl+O2----> 2Cl2+2H2O
Basis : 4 moles of HCl, Oxygen required= 1mole Air required= 1/0.21=4.762 moles
Air supplied =35% excess, Air supplied =1.35*4.762=6.43 moles
Oxygen supplied= 6.43*0.21=1.35 Nitrogen = 6.43*0.79=5.08 moles
But the Reactin is 75% complete, Hence HCl converted= 4*0.75= 3 moles, HCl remaining = 4-3=1 moles
OXygen consumed= 0.75 moles, oxygen remaining= 1.35-0.75= 0.6 moles,
Prodcut formed ::Cl2 and H2O= 2times that of oxygen =2*0.75=1.5 moles
Cl2 formed =1.5 moles and H2O formed =1.5 moles
Products : HCl = 1 mole Oxygen =0.6 moles. Nitrogen = 5.08 moles Cl2= 1.5 moles and H2O= 1.5 moles
Total= 1+0.6+5.08+3=9.68 moles
Product composition : HCl =*1/9.68=0.1033 Oxygen= 100*0.6/9.68=0.06198 Cl2 =1.5/9.68=0.1549
H2O = 0.1549 and N2=5.08/9.68=0.5248
b) Basis Still remain the same in this case also
Air supplied =1.4*4.762=6.67 moles
N2= 6.67*0.79=5.27 , Oxygem =6.67-5.27=1.4
Since the reaction is 80% complete, HCl remaining= 4*0.2= 0.8 moles
Oxygen consumed =0.8 and oxygen remaiing=1.4-0.8= 0.6 Cl2 and H2O formed each= 1.6 moles
Prodcts : HCl =0.8moles, O2= 0.6 moles N2= 5.27, H2O= 1.6moles and Cl2= 1.6 moles
Total of products= 0.8+0.6+5.27+1.6+1.6=9.87moles
Composition : HCl= 0.8/9.87=0.0811 O2= 0.6/9.87=0.0608 N2= 5.27/9.87=0.5339
H2O= 1.6/9.87=0.1621 and Cl2= 0.1621