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  4. 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 :

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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. $HCl$ ( assuming $100 \%$ conversion) is :

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Solution:

$2 H _3 AsO _4+5 H _2 S \xrightarrow{\text { Conc HCl }}AsS _5+8 H _2 O$
2 moles of Arsenic Acid $\longrightarrow 1$ mole of Arsenic Pentasulphide
1 moles of Arsenic Acid $\longrightarrow 1 / 2$ mole of Arsenic Pentasulphide
$\frac{\text { Molar mass of } H _3 AsO _4=141}{\text { Molar mass of } As _2 S _5=308}$ number of moles of $H _3 AsO _4=\frac{35.5}{141}=0.25$
$\therefore$ number of moles of $A s_2 S_5=\frac{0.25}{2}=0.125\, mol$