Q22 / 90 · JEE Main 2016 (09 Apr Online) · Chemistry

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

  • A
    0.25 mol
  • B
    0.50 mol
  • C
    0.333 mol
  • Correct answer
    D
    0.125 mol

Solution

2H3AsO4+5H2S→Conc.HClAs2S5+8H2O

2 moles of Arsenic Acid → 1 mole of Arsenic Pentasulphide

1 moles of Arsenic Acid → 1/2 mole of Arsenic Pentasulphide

Molar mass of H3AsO4=141; Molar mass of As2S5=308

∴ Number of moles of H3AsO4=35.5141=0.25

∴  Number of moles of As2S5=0.252=0.125 mol

Hence, the correct answer is option: 'D'.

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