Chemistry, 23.06.2019 07:40 Aaron5795

)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.

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Answer from: JellalFernandes

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}${O}}} = 0.1044\times 1.70 = 0.177\\\\$

Answer from: Quest

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explanation:

Answer from: Quest

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)in the deacon process for the manufacture of chlorine, hcl and o2 react to form cl2 and h2o. suffic...

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