The treatment of an aqueous solution of 3 . 74   g of Cu NO 3 2 with excess KI results in a brown…

Number of moles of $\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2=\frac{3.74}{187}=0.02$

$2\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+4\mathrm{KI}\to {\mathrm{Cu}}_2{\mathrm{I}}_2\downarrow +{\mathrm{I}}_2+4{\mathrm{KNO}}_3$

Number of moles of ${\mathrm{Cu}}_2{\mathrm{l}}_2$ precipitated $=0.01$

${\mathrm{I}}_2\left(\mathrm{brown} \mathrm{solution}\right)+{\mathrm{H}}_2\mathrm{S}\to \mathrm{S}\downarrow +2\mathrm{HI}$

Number of moles of $\mathrm{S}$ precipitated $=0.01$

Mass of $\mathrm{S}$ precipitates $=\left(0.01\times 32\right)\mathrm{g}=0.32 \mathrm{g}$