2CuSO-4 - K-4-Fe-CN-6- - A- 2K-2SO-4- A is-None of the aboveCu-2-Fe-CN-6-Cu-2-Fe-CN-4-Cu-Fe-CN-z-6-

The correct option is B 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 IX

Chemistry

Copper II Sulphate

2CuSO_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 $C u^{+ 2}$ .

Suggest Corrections

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