Question bank/JEE Main/Chemistry

The amount of arsenic pentasulphide (As_2S_5) that can be
obtained when 35.5arsenic acid (H_3AsO_4) is treated with excess H_2S
in the presence of conc. HCl (assuming 100 conversion) is

A0.333 molB0.125 molC0.25 molD0.50 mol

Show answer and solution▾

H_3AsO_4 + H_2S + HCL [IMAGE] As_2S_5
Initial moles of H_3AsO_4 = $\frac{35.5}{142}\mathbf{=}\frac{1}{4}$
mol
Applying POAC on As atoms; moles of As_2S_5$= \frac{1}{8}$ mol

Source: Grass-G / CH-05 Chemistry Paper 10 October.docx