Answer to Question #180905 in General Chemistry for Mary

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 mole% N2) is fed to provide 35% excess oxygen and the fractional conversion of

HCl is 85%. Calculate the composition (% mole) of the product stream components. (Basis: 100 moles

HCl)

Answer 1

Basis : 100 moles of HCl

Fractional conversion of HCl = 85%

HCl reacted = 85 moles

HCl unreacted = 15 moles

Oxygen is supplied in 35% excess

$2HCl+\dfrac{1}{2}O_2\longrightarrow Cl_2+H_2O$

From Balanced chemical reaction,

2 moles of HCl reacts to form 1 mole of Cl₂ and 1 mole of H₂O

From 85 moles of HCl,

Cl₂ formed = 42.5 moles

H₂O formed = 42.5 moles

2 moles of HCl reacts with $\dfrac{1}{2}$ moles of O₂

85 moles of HCl reacts with $=85\times\dfrac{1}{4}=21.25$ moles of O₂

Moles of O₂ supplied = $(\dfrac{35}{100}\times21.25)+21.25=28.68$ moles

Moles of O₂ unreacted = moles of O₂ supplied $-$ moles of O₂ reacted = 7.43 moles

100 moles of air contains 21 moles of O₂

28.68 moles of O₂ is contained in 136.57 moles of air

100 moles of air contains 79 moles of N₂

136.57 moles of air contains = 107.89 moles of N₂

Product stream contains,

N₂ = 107.89 moles = $\dfrac{107.89}{215.32}\times100=50.10\%$

O₂ = 7.43 moles = $\dfrac{7.43}{215.32}\times100=3.45\%$

HCl = 15 moles = $\dfrac{15}{215.32}\times100=6.96\%$

Cl₂ = 42.5 moles = $\dfrac{42.5}{215.32}\times100=19.73\%$

H₂O = 42.5 moles = $\dfrac{42.5}{215.32}\times100=19.73\%$