The amount of arsenic pentasulphide that can be obtained when 35 . 5 g arsenic acid is treated with excess H 2 S in the presence of conc. HCl (assuming 100 % conversion) is

Step-by-step solution

2 H 3 A s O 4 + 5 H 2 S → C o n c . H C l A s 2 S 5 + 8 H 2 O 2 moles of Arsenic Acid → 1 mole of Arsenic Pentasulphide 1 moles of Arsenic Acid → 1 / 2 mole of Arsenic Pentasulphide M o l a r   m a s s   o f   H 3 A s O 4 = 141 ;   M o l a r   m a s s   o f   A s 2 S 5 = 308 ∴ Number of moles of H 3 A s O 4 = 35.5 141 = 0.25 ∴   Number of moles of A s 2 S 5 = 0.25 2 = 0.125   m o l