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 :
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 :
Updated On: Sep 30, 2024
- 0.50 mol
- 0.25 mol
- 0.125 mol
- 0.333 mol
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The Correct Option is C
Solution and Explanation
${2H_3 AsO_4 + 5H_2 S ->[Conc\,HCl] As S_5 + 8H_2O }$
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 $As_{2}S_{5}=\frac{0.25}{2}=0.125\,mol$
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 $As_{2}S_{5}=\frac{0.25}{2}=0.125\,mol$
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