the minimum quantity in gramsof H2S needed to precipitate 63.5 g of Cu+2 will be nearly
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Answered by
104
Reaction for the given condition will be ⇒
Cu²⁺ + H₂S ------------→ Cu₂S + 2[H]⁺
In this Reaction,
1 mole of the Cu²⁺ reacts with the one mole of the H₂S.
Molar Mass of the H₂S = 2 + 32 = 34 g.
∴ 63.5 g of the Cu²⁺ reacts with the 34 g of the H₂S.
Hence, the mass of the H₂S needed in the grams is 34 g.
Hope it helps.
Cu²⁺ + H₂S ------------→ Cu₂S + 2[H]⁺
In this Reaction,
1 mole of the Cu²⁺ reacts with the one mole of the H₂S.
Molar Mass of the H₂S = 2 + 32 = 34 g.
∴ 63.5 g of the Cu²⁺ reacts with the 34 g of the H₂S.
Hence, the mass of the H₂S needed in the grams is 34 g.
Hope it helps.
Answered by
14
H2S needed is 34 grams.............
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