MCQ
The amount of arsenic pentasulphide that can be obtained when $35.5\,g$ arsenic acid is treated with excess $H_2S$ in the presence of cone. $HCl$ (assuming $100\%$ conversion) is $....\,\,mol$
- A$0.25$
- B$0.50$
- C$0.333$
- ✓$0.125$
✓
Answer
Correct option: D.
$0.125$
d
$\underset{1\,moles}{\mathop{\underset{2\,moles\,}{\mathop{2{{H}_{3}}As{{O}_{4}}}}\,}}\,+5{{H}_{2}}S\xrightarrow{Conc.HCl}\underset{1/2\,moles}{\mathop{\underset{1\,mole}{\mathop{A{{s}_{2}}{{S}_{5}}}}\,}}\,+8{{H}_{2}}O$
$\underset{1\,moles}{\mathop{\underset{2\,moles\,}{\mathop{2{{H}_{3}}As{{O}_{4}}}}\,}}\,+5{{H}_{2}}S\xrightarrow{Conc.HCl}\underset{1/2\,moles}{\mathop{\underset{1\,mole}{\mathop{A{{s}_{2}}{{S}_{5}}}}\,}}\,+8{{H}_{2}}O$
$\therefore $ number of moles of $H_3AsO_4$
$=\frac{35.5}{142}=0.25$
$\therefore $ number of moles of $As_2S_5$
$=\frac{0.25}{2}=0.125\,mol.$
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