Iodine is formed when KI reacts with a solution of :

Z n S O_{4}

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C u S O_{4}

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F e S O_{4}

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(N H_{4})_{2} S O_{4}

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Solution

The correct option is B $C u S O_{4}$
Iodine is formed when KI reacts with a solution of $C u S O_{4}$
$C u S O_{4} + 2 K I \to C u I_{2} + K_{2} S O_{4}$
$2 C u I_{2} \to C u_{2} I_{2} + I_{2}$
Hence option B is correct.

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Similar questions

Q. Iodine is formed when potassium iodide reacts with a solution of:

Q. Zinc reacts with

C u {S O}_{4}

according top the equation

Z n + C u {S O}_{4} ⟶ Z n {S O}_{4} + C u

. if excess of zinc is added to

100 m l

0.05 M

C u {S O}_{4}

the amount of copper formed is:

I- ion react with iodine in aqueous solution to form I3- ion .When 1L of solution containing 1mole of KI and 0.25 mol of I2 was reacted with excess of AgNO3 ,0.80 mol of yellow precipitate was formed.The stability constant of. I3- ion is

Q. Consider the following reactions :

C u S O_{4} + F e \to F e S O_{4} + C u

F e S O_{4} + Z n \to Z n S O_{4} + F e

Among these:

Q. Zinc reacts with

{C u S O}_{4}

according to the equation,

Z n + {C u S O}_{4} \to {Z n S O}_{4} + C u

. If an excess of zinc is added to 100ml, 0.05M

{C u S O}_{4}

, the amount of copper formed in moles will be :

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Iodine is formed when KI reacts with a solution of :

The correct option is B CuSO4Iodine is formed when KI reacts with a solution of CuSO4CuSO4+2KI→CuI2+K2SO42CuI2→Cu2I2+I2Hence option B is correct.

The correct option is B $C u S O_{4}$
Iodine is formed when KI reacts with a solution of $C u S O_{4}$
$C u S O_{4} + 2 K I \to C u I_{2} + K_{2} S O_{4}$
$2 C u I_{2} \to C u_{2} I_{2} + I_{2}$
Hence option B is correct.