Question

A gas $X$ occupies $22.4$ L at S.T.P. The gas turns acidified $K_{2}C{r}_2{O}_7$ green and decolourizes acidified $KMn{O}_4$ solution. White turbidity appears when $H_{2}S$ gas is passed through aqueous solution of gas, then $X$ will be :

• A. $H_{2}S$
• B. $N{O}_2$
• C. $S{O}_2$
• D. none of these

Answer 1

Answer: (C)

$S{O}_2$

$S{O}_2$ when passed through $K_{2}C{r}_2{O}_7$ solution , it turns green.

$3S{O}_2+C{r}_2{O}_7^{2-}+2H^{+}⟶3S{O}_4^{2-}+2C{r}^{3+}+H_{2}O$

Potassium permanganate has a purple colour. Sulphur dioxide is a reducing agent. Potassium permanganate is an oxidizing agent. When sulphur dioxide reacts with potassium permanganate thesolution decolourizes. Color changes from Purple to transparent (colourless).
Reaction:
$2KMn{O}_4+5S{O}_2+2H_{2}O\to 2MnS{O}_4+2KHS{O}_4+H_{2}S{O}_4$