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% excess oxygen, and the fractional conversion of HCl is 85%. Calculate the mole fractions of the product stream components.

Answer 1

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

$\rm 4HCl + O _{2} \longrightarrow \, 2Cl _{2} + 2H _{2} O$

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

$\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}$

3. Add 35% excess

$\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}$

4. Calculate the theoretical volume of nitrogen

$\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}$

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

           4HCl     +       O₂    +    N₂   →    2Cl₂   +   2H₂O

I/L:          4               1.35         5.08         0              0

C/L:  -0.85(4)        -0.85(1)        0      +0.85(2)   +0.85(2)

E/L:     0.60             0.50        5.08       1.70          1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

$\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H _{2} {O}}} = 0.1044\times 1.70 = 0.177$