Question

Compound $A$ and $B$ are treated with $dil. HCl$ separately. The gases liberated are $Y$ and $Z$ respectively. $Y$ turns acidified dichromate paper green while $Z$ turns lead acetate paper black. The compound $A$ and $B$ are respectively.

• A. $Na_2CO_3$ and $NaCl$
• B. $Na_2SO_3$ and $Na_2S$
• C. $Na_2S$ and $Na_2SO_3$
• D. $Na_2SO_3$ and $Na_2SO_4$

Answer 1

Answer: (B)

Gas $(Y) S O_{2}$ which can be obtained from $Na _{2} SO _{3}(A)$ and gas (Z) is $H_{2} S$ which can be obtained from $N a_{2} S(B)$.
This can be easily understood by the following reactions.
$\underset{(A)}{Na _{2} SO _{3}}+2 HCl \longrightarrow 2 NaCl + \underset{(Y)}{SO _{2}} H _{2} O$
$\underset{(B)}{Na _{2}S}+2 HCl \longrightarrow 2 NaCl + \underset{(Z)}{H _{2} O}$
$K _{2} Cr _{2} O _{7}+ H _{2} SO _{4}+3 SO _{2} \longrightarrow K _{2} SO _{4}+ \underset{\text{green}}{Cr _{2}( SO _{4})_{3}}+ H _{2} O$
$Pb \left( CH _{3} COO \right)_{2}+ H _{2} S \longrightarrow \underset{\text { black }}{ PbS } \downarrow+2 CH _{3} COOH$