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. 0.125 mol
• C. 0.333 mol
• D. 0.25 mol

Answer 1

The correct option is B 0.125 mol

Arsenic acid has a molar mass of 142 g/mol. 35.5 g of arsenic acid corresponds to

$\frac{35.5}{142}=0.25$ moles.

$2H_{3}As{O}_4\underset{HCl}{\overset{H_{2}S}{\to }}A{s}_2{S}_5$

The amount of arsenic pentasulfide that can be obtained is

$\frac{0.25}{2}=0.125$ mol

Hence, the correct option is $B$