The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H₂S in the presence of cone. HCl (assuming 100% conversion) is:

 Molar mass of arsenic acid ${\mathrm{H}}_3{\mathrm{AsO}}_4\text{,}$is $M=(3+74.92+4\times 16)\mathrm{g}{\mathrm{mol}}^{-1}=143.42\mathrm{g}{\mathrm{mol}}^{-1}$

Arsenic pentasulphide is ${\mathrm{As}}_2 {\mathrm{S}}_5$ .

2 mol of${\mathrm{H}}_3{\mathrm{AsO}}_4$ will provide 1 mol of ${\mathrm{As}}_2 {\mathrm{S}}_5$ .

$35.5\mathrm{g}(={(35.5/143.92)\mathrm{mol})}_4$ of ${\mathrm{H}}_3{\mathrm{AsO}}_4$will provide $\frac{1}{2}\left(\frac{35.5}{143.92}\right)\mathrm{mol}\text{ of }{\mathrm{As}}_2{\mathrm{O}}_5$

which is $0.123 \mathrm{mol}$