Answer:
Here's what I get.
Explanation:
1. Write the chemical equation
[tex]\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O[/tex]
Assume that we start with 4 L of HCl
2. Calculate the theoretical volume of oxygen
[tex]\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}}[/tex]
3. Add 35% excess
[tex]\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}[/tex]
4. Calculate the theoretical volume of nitrogen
[tex]\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}}[/tex]
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: Â Â Â Â Â 4 Â Â Â Â Â Â Â 1.35 Â Â Â Â 5.08 Â Â Â Â 0 Â Â Â Â Â Â Â 0
C/L: Â -0.85(4) Â Â Â Â -0.85(1) Â Â Â Â 0 Â Â Â +0.85(2) Â +0.85(2)
E/L: Â Â 0.60 Â Â Â Â Â Â 0.50 Â Â Â Â 5.08 Â Â Â 1.70 Â Â Â Â Â 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
[tex]\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\\\\[/tex]
