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Question

The minimum quantity in grams of H2S needed to precipitate 63.5 g of Cu2+ ion will be:
(molar mass of Cu=63.5 g/mol)

Cu2++H2SCuS+2H+

A
63.5 g
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B
41.7 g
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C
34.0 g
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D
22.4 g
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Solution

The correct option is C 34.0 g
Cu2++H2SCuS+H2(g)

63.5 g of Cu2+ = 1 mol of Cu2+
1 mol of H2S is required to precipitate 1 mol of Cu2+
So, 1 mol H2S = 1 mol×34 g/mol=34 g.

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