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Question:

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% yield)?

0.125 mol

0.50 mol

0.333 mol

0.25 mol

Solution:

Arsenic acid has a molar mass of 142 g/mol. 35.5 g of arsenic acid corresponds to 35.5/142 = 0.25 moles.

2H3AsO4 + 5H2S → As2S5 + 8H2O

The amount of arsenic pentasulfide that can be obtained is 0.25/2 = 0.125 mol

Hence, the correct option is 0.125 mol