Yellow ppt of (a)_____ is formed when `CoCl_(2)` reacts with excess of `KNO_(2)` in presence of `CH_(3)COOH`.

Answer

To answer the question regarding the formation of a yellow precipitate when CoCl₂ reacts with excess KNO₂ in the presence of CH₃COOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Reactants**: The reactants in this reaction are cobalt(II) chloride (CoCl₂), potassium nitrite (KNO₂), and acetic acid (CH₃COOH).

2. **Understand the Reaction Conditions**: The presence of acetic acid (CH₃COOH) indicates that the reaction may occur in an acidic medium, which can influence the formation of certain products.

3. **Write the Reaction**: When CoCl₂ reacts with KNO₂ in the presence of CH₃COOH, the cobalt(II) ions react with nitrite ions to form a complex. The reaction can be written as:
\[ CoCl_2 + 6KNO_2 + 2CH_3COOH \rightarrow K_3[Co(NO_2)_6] + 2KCl + 2CH_3COOK + H_2 \]

4. **Identify the Product**: The product formed is a complex compound known as potassium hexanitrocobaltate(II), represented as K₃[Co(NO₂)₆]. This compound is known to form a yellow precipitate.

5. **Conclusion**: Therefore, the yellow precipitate formed in this reaction is K₃[Co(NO₂)₆].

### Final Answer:
The yellow precipitate of **K₃[Co(NO₂)₆]** is formed when CoCl₂ reacts with excess KNO₂ in the presence of CH₃COOH.

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