$1\mathrm{.}5.$ In the manufacture of vinyl chloride$($VC$)$ by the pyrolysis of dichloroethane $($DCE$),$ the reactor conversion is limited to $55\%$ to reduce carbon formation, which fouls the reactor tubes.

Calculate the quantitiy of DCE fed to the reactor to produce $5000k\frac{g}{h}$ of VC$.$

Solution

Basis: $5000k\frac{g}{h}$ VC $($the required quantity$).$

Reaction: $C_2{H}_{4}C{l}_2{C}_2{H}_{3}Cl+HCl$

mol weights DCE $99,VC62\mathrm{.}5$

kmo$\frac{l}{h}VC$ produced $=\frac{5000}{62\mathrm{.}5}={80}_{{?}_{?}}$

From the stoichiometric equation, $1$kmol DCE produces $1$kmol VC$.$ Let $x$ be DCE feed kmo$\frac{l}{h}$ :

Per cent conversion $=55=\frac{80}{x}100$

$x=\frac{\mathrm{.}80}{0\mathrm{.}55}=145\mathrm{.}5$kmo$\frac{l}{h_{{?}_{?}}}$

In this example the small loss of DCE to carbon and other products has been neglected. All the DCE reacted has been assumed to be converted to VC$.$

$32$