The amount of arsenic pentasulphide that can | vedclass

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(D) The balanced chemical equation is: $2H_3AsO_4 + 5H_2S \xrightarrow{	ext{conc. } HCl} As_2S_5 + 8H_2O$ First,calculate the molar mass of arsenic a

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(D) The balanced chemical equation is: $2H_3AsO_4 + 5H_2S \xrightarrow{	ext{conc. } HCl} As_2S_5 + 8H_2O$ First,calculate the molar mass of arsenic acid $(H_3AsO_4)$: $M = (3 	imes 1) + 74.92 + (4 	imes 16) \approx 142 \ g/mol$. Number of moles of $H_3AsO_4 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{35.5 \ g}{142 \ g/mol} = 0.25 \ mol$. According to the stoichiometry of the reaction,$2 \ mol$ of $H_3AsO_4$ produce $1 \ mol$ of $As_2S_5$. Therefore,$0.25 \ mol$ of $H_3AsO_4$ will produce: $\frac{0.25}{2} = 0.125 \ mol$ of $As_2S_5$.

The amount of arsenic pentasulphide that can be obtained when $35.5 \ g$ of arsenic acid $(H_3AsO_4)$ is treated with excess $H_2S$ in the presence of conc. $HCl$ (assuming $100\%$ conversion) is $.... \ mol$.

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