A white crystalline salt A reacts with dilute | vedclass

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(B) Gas $B$ turns the colour of acidified $K_{2}Cr_{2}O_{7}$ green,thus it is $SO_{2}$. $SO_{2}$ is obtained along with a yellow precipitate of sulfur

Vedclass · Accepted Answer

$Na_{2}S_{2}O_{3}, SO_{2}, Cr_{2}(SO_{4})_{3}$

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(B) Gas $B$ turns the colour of acidified $K_{2}Cr_{2}O_{7}$ green,thus it is $SO_{2}$. $SO_{2}$ is obtained along with a yellow precipitate of sulfur when thiosulfate is treated with dilute acids. Thus,$A$ is $Na_{2}S_{2}O_{3}$,$B$ is $SO_{2}$,and $C$ is $Cr_{2}(SO_{4})_{3}$.The reactions are as follows:$Na_{2}S_{2}O_{3} + 2HCl ightarrow 2NaCl + H_{2}O + SO_{2} + S 	ext{ (yellow precipitate)}$$K_{2}Cr_{2}O_{7} + 3SO_{2} + H_{2}SO_{4} ightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} 	ext{ (green solution)} + H_{2}O$

$A$ white crystalline salt $A$ reacts with dilute $HCl$ to liberate a suffocating gas $B$ and also forms a yellow precipitate. The gas $B$ turns potassium dichromate acidified with dilute $H_{2}SO_{4}$ to a green coloured solution $C$. $A$,$B$ and $C$ are respectively

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