Question

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess ${H_2S}$ in the presence of conc. HCl ( assuming 100% conversion) is :

• A. 0.50 mol
• B. 0.25 mol
• C. 0.125 mol
• D. 0.333 mol

Answer 1

Answer: (C)

0.125 mol

Answer 2

${2H_3 AsO_4 + 5H_2 S ->[Conc\,HCl] As S_5 + 8H_2O }$
2 moles of Arsenic Acid ${->}$ 1 mole of Arsenic Pentasulphide
1 moles of Arsenic Acid ${->}$ 1/2 mole of Arsenic Pentasulphide
$\frac{Molar\,mass\, of\, H_{3}AsO_{4}=141}{Molar\,mass\,of\, As_{2}S_{5}=308} \,$number of moles of $H_{3}AsO_{4}=\frac{35.5}{141}=0.25$
$\therefore$ number of moles of $As_{2}S_{5}=\frac{0.25}{2}=0.125\,mol$