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  • The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :<div class='

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :

Option: 1

0.50 mol


Option: 2

0.25 mol


Option: 3

0.125 mol


Option: 4

0.333 mol


Answers (1)

best_answer

Question is saying we have to find the amount of arsenic pentasulphide obtained when 35.5g arsenic acid is treated with excess with H2S in the presence of Conc. HCl.

moles of arsenic acid = weight / molar mass = 35.5/142 = 0.25 moles

then write the chemical reaction equation-

The chemical equation is as follows:
\mathrm{2H_{3}AsO_{4}}\overset{H_{2}S/HCl}{\rightarrow}\, As_{2}S_{5}

Thus 2 moles of arsenic acid produces one mole of arsenic pentasulphide.

This means one mole of arsenic pentasulphide is forming from  2 moles of arsenic acid.

now, 1/2 mole of arsenic pentasulphide is forming from  1 mole of arsenic acid.

now, 0.25/2 mole of arsenic pentasulphide is forming from  0.25 moles of arsenic acid.

Thus, moles of arsenic pentasulphide produced

= 0.25 / 2
= 0.125 moles

Therefore, Option(3) is correct 

Posted by

Ajit Kumar Dubey

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