A mixture of 50.0g of S and 100.0g ${\mathrm{Cl}}_2$ reacts of form ${\mathrm{S}}_2{\mathrm{Cl}}_2$  The mass of ${\mathrm{S}}_2{\mathrm{Cl}}_2$ formed will be

The reaction is $$\begin{gathered} 2\mathrm{S}+{\mathrm{Cl}}_2\to {\mathrm{S}}_2{\mathrm{Cl}}_2 \\ {n}_{0}50.0\mathrm{g}100.0\mathrm{g} \end{gathered}$$

$n_0\left(\mathrm{S}\right)=\frac{50.0\mathrm{g}}{32\mathrm{g}{\mathrm{mol}}^{-1}}=1.5625\mathrm{mol}$

$n_0\left({\mathrm{Cl}}_2\right)=\frac{100.0\mathrm{g}}{71.0\mathrm{g}{\mathrm{mol}}^{-1}}=1.4084\mathrm{mol}$

${\xi }_{max}\left(\mathrm{S}\right)=\frac{n_0\left(\mathrm{S}\right)}{v\left(\mathrm{S}\right)}=\frac{1.5625\mathrm{mol}}{2}=0.7812\mathrm{mol}$${\xi }_{max}\left({\mathrm{Cl}}_2\right)=\frac{n_0\left({\mathrm{Cl}}_2\right)}{v\left({\mathrm{Cl}}_2\right)}=\frac{1.4084\mathrm{mol}}{1}=1.4084\mathrm{mol}$S acts as limiting reagent. Hence

$\begin{array}{r}\mathrm{\Delta }n\left({\mathrm{S}}_2{\mathrm{Cl}}_2\right)=v\left({\mathrm{S}}_2{\mathrm{Cl}}_2\right){\xi }_{max}\left(\mathrm{S}\right)=\left(1\right)\left(0.7812\mathrm{mol}\right)=0.7812\mathrm{mol}\\ \mathrm{\Delta }m\left({\mathrm{S}}_2{\mathrm{Cl}}_2\right)=\mathrm{\Delta }n\left({\mathrm{S}}_2{\mathrm{Cl}}_2\right)M\left({\mathrm{S}}_2{\mathrm{Cl}}_2\right)=\left(0.7812\mathrm{mol}\right)\left(135\mathrm{g}{\mathrm{mol}}^{-1}\right)=105.5\mathrm{g}\end{array}$