Question

21. In the Deacon Process to make chlorine, HCl and O₂ react to form Cl₂ and H₂O. Sufficient air (21 mole% O₂, 79% N₂) is fed to allow 25% excess oxygen and the final conversion of HCl is 70%. Calculate the mole fractions of all components in the product stream.

Answer 1

Answer:
4HCl +O₂ ? 2Cl₂ + 2H₂O
From the balanced equation 4 moles of HCl reacts with 1 mole of O₂ to form 2 moles of Cl₂ and 2 moles of H₂O.
Gram molecular weight of 1 mole of HCl = 36.5 grams
Gram molecular weight of 4 moles of HCl = (4*36.5 grams) = 146 gram
146 grams of HCl combines with 32 grams of O₂ to form products.
Final conversion is 70% of HCl = (70*146)/100 = 102.2 grams
146 grams of HCl combines with 32 grams of O2.
102.2 grams of HCl combines with = (102.2*32)/146
= 22.4 grams
Number of moles of HCl = 102.2/146 = 0.7
Number of moles of O2 = 22.4/32 = 0.7
Thus 0.7 moles of HCl combines with 0.7 moles of O2 to form 0.35 moles of Cl₂ and H₂O.
Total number of moles = 0.7 +0.7 +0.35 +0.35
= 2.1
Molefraction of HCl = number of moles of HCl/Total number of moles
= 0.7/2.1
= 0.33
Molefraction of O₂ = number of moles of O₂/Total number of moles
= 0.7/2.1
= 0.33
Molefraction of Cl₂ = number of moles of Cl₂/Total number of moles
= 0.35/2.1
= 0.167
Molefraction of H₂O = number of moles of H₂O/Total number of moles
= 0.35/2.1
= 0.167