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Pb(CH(3)COO)(2) gives . . . Colour with ...

`Pb(CH_(3)COO)_(2)` gives . . . Colour with `H_(2)S`

A

Orange

B

Red

C

Black

D

White

Text Solution

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The correct Answer is:
To solve the question regarding the reaction of lead acetate (Pb(CH₃COO)₂) with hydrogen sulfide (H₂S), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: - The reactant is lead acetate, which has the formula Pb(CH₃COO)₂. - The other reactant is hydrogen sulfide, which has the formula H₂S. 2. **Write the Reaction**: - When lead acetate reacts with hydrogen sulfide, the following reaction occurs: \[ Pb(CH₃COO)₂ + H₂S \rightarrow PbS + 2 CH₃COOH \] - Here, lead acetate reacts with hydrogen sulfide to produce lead sulfide (PbS) and acetic acid (CH₃COOH). 3. **Identify the Products**: - The products of the reaction are: - Lead sulfide (PbS) - Acetic acid (CH₃COOH) 4. **Determine the Color of the Precipitate**: - Lead sulfide (PbS) is known to form a black precipitate when it is produced in a reaction. 5. **Conclusion**: - Therefore, the color of the precipitate formed when lead acetate reacts with hydrogen sulfide is black. ### Final Answer: The color of the precipitate formed when Pb(CH₃COO)₂ reacts with H₂S is **black**. ---
To solve the question regarding the reaction of lead acetate (Pb(CH₃COO)₂) with hydrogen sulfide (H₂S), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: - The reactant is lead acetate, which has the formula Pb(CH₃COO)₂. - The other reactant is hydrogen sulfide, which has the formula H₂S. ...
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