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 :

Question

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 : (A) 0.50 mol (B) 0.25 mol (C) 0.125 mol (D) 0.333 mol

Tardigrade Admin · Accepted Answer

2 H 3 AsO 4+5 H 2 S xrightarrow text Conc HCl AsS 5+8 H 2 O 2 moles of Arsenic Acid longrightarrow 1 mole of Arsenic Pentasulphide 1 moles of Arsenic Acid longrightarrow 1 / 2 mole of Arsenic Pentasulphide ( text Molar mass of H 3 AsO 4=141/ text Molar mass of As 2 S 5=308) number of moles of H 3 AsO 4=(35.5/141)=0.25 ∴ number of moles of A s2 S5=(0.25/2)=0.125 mol

Q. The amount of arsenic pentasulphide that can be obtained when 35.5g arsenic acid is treated with excess H2​S in the presence of conc. HCl ( assuming 100% conversion) is :

0.50 mol

0.25 mol

0.125 mol

0.333 mol

Q. 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. $H Cl$ ( assuming $100%$ conversion) is :