JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 21)
The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :
0.50 mol
0.25 mol
0.125 mol
0.333 mol
Explanation
2H3 As O4 + 5H2S $$\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{HCl}^{conc}} $$ As2S5 + 8H2O
35.5 g of H3AsO4 = $${{35.5} \over {142}}$$ = 0.25 moles
Let, As2S5 produced = n moles.
$$\therefore\,\,\,$$ $${{0.25} \over 2} = {n \over 1}$$
$$ \Rightarrow $$$$\,\,\,$$ n = 0.125 mol.
35.5 g of H3AsO4 = $${{35.5} \over {142}}$$ = 0.25 moles
Let, As2S5 produced = n moles.
$$\therefore\,\,\,$$ $${{0.25} \over 2} = {n \over 1}$$
$$ \Rightarrow $$$$\,\,\,$$ n = 0.125 mol.
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