${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{HCl}}_{\left(\mathrm{dry}\right)}\frac{\text{ anhydrous }}{{\mathrm{ZnCl}}_2}\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}$ In this reaction, the function of ZnCl₂ is

to prevent the backward reaction

to help the formation of Cl^–

to speed up the formation of C₂H₅Cl

to form C2H₅ ⁽⁺⁾ from C₂H₅OH