`A overset(KCN"added slowly")(to) underset(("Reddish brown precipitate"))(B) overset("In excess KCN precipitate dissolves")(to)underset(("Yellow solution"))(C)`
`A overset(K_(4)Fe(CN)_(6))(to) underset(("Intense blue precipitate"))(D) overset(NaOH)(to) ("Red precipitate")`
`A overset(Na_(2)HPO_(4))(to) underset(("Yellowish white precipitate"))(E)`
How many produces are correctly match?
`A=FeCl_(3), A=FeCl_(2), A=CuCl_(2), B=Fe(CN)_(3)`
`B=K_(3)Fe(CN)_(6), C=K_(3)Fe(CN)_(6), C=K_(4)[Fe(CN)_(6)], D=Fe_(4)[Fe(CN)_(6)]_(3)`
`D=Fe[Fe(CN)_(6)], E=FePO_(4)`

Explanation

To solve the problem, we need to analyze the reactions involving compound A and determine the products formed in each case. We will then match these products with the given options.

### Step-by-Step Solution:

1. **Identify Compound A**:
- We start with the assumption that A could be FeCl₃ (ferric chloride).
- When KCN is added slowly to FeCl₃, it forms a reddish-brown precipitate (B). This precipitate is likely ferric cyanide, which is consistent with the reaction.

2. **Excess KCN**:
- When excess KCN is added to the reddish-brown precipitate formed from FeCl₃, the precipitate dissolves to form a yellow solution (C). The soluble complex formed here is K₃[Fe(CN)₆], which is indeed yellow.

3. **Reaction with K₄[Fe(CN)₆]**:
- Next, when FeCl₃ reacts with K₄[Fe(CN)₆], it produces an intense blue precipitate (D). This precipitate is known as Prussian blue, which has the formula Fe₄[Fe(CN)₆]₃.

4. **Reaction with NaOH**:
- When NaOH is added to the intense blue precipitate (D), it forms a red precipitate. This red precipitate is ferric hydroxide (Fe(OH)₃).

5. **Reaction with Na₂HPO₄**:
- Finally, when FeCl₃ reacts with Na₂HPO₄, it produces a yellowish-white precipitate (E), which is ferric phosphate (FePO₄).

### Summary of Products:
- A = FeCl₃
- B = Reddish-brown precipitate (Ferric cyanide)
- C = Yellow solution (K₃[Fe(CN)₆])
- D = Intense blue precipitate (Fe₄[Fe(CN)₆]₃)
- E = Yellowish-white precipitate (FePO₄)

### Matching with Given Options:
- A = FeCl₃ (Correct)
- B = K₃[Fe(CN)₆] (Correct)
- C = K₄[Fe(CN)₆] (Correct)
- D = Fe₄[Fe(CN)₆]₃ (Correct)
- E = FePO₄ (Correct)

### Conclusion:
All five products are correctly matched.

### Final Answer:
**Total number of correctly matched products = 5**

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