In the reaction, 

$x{\mathrm{FeCl}}_3+y{\mathrm{H}}_2\mathrm{S}⟶{\mathrm{FeCl}}_2+\mathrm{S}+\mathrm{HCl}$

Mass balance and Charge Balance :

Remove the spectator ion - ${\mathrm{Cl}}^{-}$

${\mathrm{Fe}}^{3+}+{\mathrm{H}}_2\mathrm{S}\to {\mathrm{Fe}}^{2+}+\mathrm{S}+\mathrm{H}{\mathrm{Cl}}^2$

Oxidation Half :

${\mathrm{Fe}}^{3+}+1e^{-}\to {\mathrm{Fe}}^{2+}\mid$

Reduction Half:

${\mathrm{H}}_2\mathrm{S}\to S+2{\mathrm{H}}^{+}+2e^{-}$

Total loss of electrons = total gain of electrons

$2{\mathrm{Fe}}^{3+}+{\mathrm{H}}_2\mathrm{S}\to 2{\mathrm{Fe}}^{2+}+\mathrm{S}+2{\mathrm{H}}^{+}$

Add the spectator ion – ${\mathrm{Cl}}^{-}$

$2{\mathrm{FeCl}}_3+{\mathrm{H}}_2\mathrm{S}\to 2{\mathrm{FeCl}}_2+\mathrm{S}+2\mathrm{HCl}$

${\mathrm{FeCl}}_3$ acts as an oxidation agent

So, In the reaction : $x{\mathrm{FeCl}}_3+y{\mathrm{H}}_2\mathrm{S}\to {\mathrm{FeCl}}_2+\mathrm{S}+\mathrm{HCl}$

value of x =2 and y =1.