To manufacture chlorine, $HCl$ and $O_2$ react to form $C{l}_2$ and $H_{2}O.$ To drive this reaction and

achieve $85\%$ fractional conversion of $HCl,$ sufficient air $(21\%O_2,79\%N_2)$ is fed into the plant, at

$35\%$ excess oxygen.

a$)$ Use molecular species balance or atomic species balance to calculate the mole fractions $($Y$)$

of all chemical components $($i$.$e$.,HCl,O_2,N_{2}C{l}_2,H_{2}O).$

Hints: Begin by balancing the chemical reaction:

$HCl+$

$O_2$

$C{l}_2+$

$n_{\text{t}\text{o}\text{t}\text{,},\text{o}\text{u}\text{t}}={\displaystyle \underset{?}{\overset{?}{}}}{n}_j=n_{HCl}+n_{O2}+n_{N2}+n_{C2}+n_{H2O}$

b$)$ Using the extent of reaction method instead, determine the molar output for each species.

You note that an alternative to using air as an $O_2$ source would be to feed pure $O_2$ to the

plant reactor.

c$)$ Speculate on potential costs $/$ savings to justify whether it's worth using pure $O_2$ for the

process.