MHT CET Chemistry question
The minimum quantity of needed to precipitate 63.5 g of will be nearly:
Answer: (C)
Solution
Cu²⁺ + H₂S → CuS + 2H⁺, so 63.5 g (1 mol) Cu²⁺ needs 1 mol H₂S = 34 g; 63.5 g,
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