JEE Main 2016 · 9 Apr · Shift 1 · Chemistry Q31

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ChemistrySingle correct+4 / -1Question 31 of 90

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

0.50 mol

0.25 mol

0.125 mol

0.333 mol

Answer and worked solution

Official answer: C

$\ce{H3AsO4}$ is 142 g mol⁻¹, so 35.5 g is 0.25 mol of As. $\ce{As2S5}$ contains two As, so 0.125 mol forms.

Topics:Stoichiometry Mole concept

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JEE Main 2016 · 9 Apr · Shift 1, Chemistry question 31