2 CuSO 4- K 4- Fe CN 6- - A -2 K 2 SO 4- A isA- Cu 2- Fe CN 6-B- Cu - Fe CN z -6-C- Cu 2- Fe CN 4-D- None of the above

The correct option is A Cu_2[Fe(CN)_6]The reaction can be given as: 2CuSO_4 + K_4[Fe(CN)_6] → Cu_2Fe(CN)_6+ 2K_2SO_4. The product formed Cu_2[Fe(CN)_6] is cal ...

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Byju's Answer

Standard X

Chemistry

Roasting

2 CuSO 4+ K 4...

Question

$2 C u S O_{4} + K_{4} [F e (C N)_{6}] ⟶ A + 2 K_{2} S O_{4} . A$ is

C u_{2} [F e (C N)_{6}]

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C u [F e (C N) z - 6]

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C u_{2} [F e (C N)_{4}]

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None of the above

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Solution

The correct option is A $C u_{2} [F e (C N)_{6}]$
The reaction can be given as: $2 C u S O_{4} + K_{4} [F e (C N)_{6}] \to C u_{2} F e (C N)_{6} + 2 K_{2} S O_{4}$ .
The product formed $C u_{2} [F e (C N)_{6}]$ is called cupric ferrocyanide. The product obtained is reddish brown precipitate and this reaction is used as a test for $C u^{+ 2}$ .

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Similar questions

2 C u S O_{4} + K_{4} [F e (C N)_{6}] ⟶ A + 2 K_{2} S O_{4} . A

2 C u S O_{4} + K_{4} [F e (C N)_{6}] ⟶ A + 2 K_{2} S O_{4} . A

Q. With

C u^{2 +}

ions

[F e (C N)_{6}]^{4 -}

gives

C u_{2} [F e (C N)_{6}]

(Cupric ferrocyanide). The colour of

C u_{2} [F e (C N)_{6}]

is?

Q. Mineral (A)

d i l . H_{2} S O_{4} ⟶

(B) + (C) + (D) (gas)
(D) +

C H_{3} C O O)_{2} P b △ ⟶

black ppt

(B) + (C) (D) \to

E (black ppt) separated from (C) by filtration
(E) +

H N O_{3} d i l . △ \to

(F) blue coloured solution
(F) +

K_{4} [F e (C N)_{6}] ⟶

K_{3} [F e (C N)_{6}] ⟶

(H) blue

A to H are identified as: (A) :

C u F e S_{2}

, (B) :

C u S O_{4}

F e S O_{4}

(D) :

H_{2} S

(E) :

C u S

(F) :

C u (N O_{3})_{2}

(G) :

C u_{2} [F e (C N)_{6}]

(H) :

K F e^{I I} [F e^{I I I} (C N)_{6}]

, Turnbull's blue

Show the above reaction.

K_{4} [F e (C N)_{6}]

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2CuSO4+K4[Fe(CN)6]⟶A+2K2SO4.A is

The correct option is A Cu2[Fe(CN)6]The reaction can be given as: 2CuSO4+K4[Fe(CN)6]→Cu2Fe(CN)6+2K2SO4. The product formed Cu2[Fe(CN)6] is called cupric ferrocyanide. The product obtained is reddish brown precipitate and this reaction is used as a test for Cu+2.

$2 C u S O_{4} + K_{4} [F e (C N)_{6}] ⟶ A + 2 K_{2} S O_{4} . A$ is