A white powder $A$ on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dilute $H C l$ and the solution gives a white precipitate with $K_{4} [F e (C N)_{6}]$ . Compound $A$ is:

C a C O_{3}

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Z n C O_{3}

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C a S O_{3}

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C u C O_{3}

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Solution

The correct option is B $Z n C O_{3}$
$A$ is $Z n C O_{3}$
The reaction take place as follows:

$Z n C O_{3} \to Z n O$ (white residue) $+ C O_{2}$ , since $C O_{2}$ is non-combustible gas.

$Z n O$ (white) $\to Z n^{2 +}$ (yellow) $+ 2 e^{-}$

$Z n O +$ dil. $H C l \to Z n C l_{2} (a q) + H_{2} O$

$2 Z n C l_{2} (a q) + K_{4} [F e (C N)_{6}] \to Z n_{2} [F e (C N)_{6}]$ (white ppt.) $+ 4 K C l$

Option B is correct.

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Similar questions

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A

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A white powder A on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dilute HCl and the solution gives a white precipitate with K4[Fe(CN)6]. Compound A is:

The correct option is B ZnCO3A is ZnCO3The reaction take place as follows:ZnCO3→ZnO (white residue) +CO2 , since CO2 is non-combustible gas.ZnO(white) →Zn2+ (yellow) +2e−ZnO + dil. HCl→ZnCl2(aq)+H2O 2ZnCl2(aq)+K4[Fe(CN)6]→Zn2[Fe(CN)6] (white ppt.) +4KClOption B is correct.

A white powder $A$ on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dilute $H C l$ and the solution gives a white precipitate with $K_{4} [F e (C N)_{6}]$ . Compound $A$ is: