Question

A mixture of three gases $A,B$ and $C$ is passed first into acidified $K_{2}C{r}_2{O}_7$ solution when $A$ is absorbed turning the solution green. The remainder of the gas is passed through excess of lime water which turns milky resulting in the absorption of $B$. The residual gas $C$ is absorbed by alkaline pyrogallol solution. However the original mixture does not turn lead acetate paper black. Identify $A,B$ & $C$:

• A. $\left(A\right)-S{O}_2$$\left(B\right)-CO$$\left(C\right)-O_2$
• B. $\left(A\right)-S{O}_3$$\left(B\right)-CO$$\left(C\right)-C{O}_2$
• C. $\left(A\right)-S{O}_2$$\left(B\right)-C{O}_2$$\left(C\right)-O_2$
• D. $\left(A\right)-S{O}_2$$\left(B\right)-CO$$\left(C\right)-C{O}_2$

Answer 1

The correct option is C $\left(A\right)-S{O}_2$$\left(B\right)-C{O}_2$$\left(C\right)-O_2$

Option $\left(A\right)$ is correct.

$S{O}_2$ gas turns acidified $K_{2}C{r}_2{O}_7$ solution turns green. Hence, the gas A is $S{O}_2$

$C{O}_2$ gas turns lime water milky. Hence, the gas $B$ is $C{O}_2$

$O_2$ gas is absorbed by alkaline pyrogallol solution and turns it into dark color ortho quinone. Hence the gas $C$ is $O_2$.
Since, the lead acetate paper is not turned black, this indicates absence of $H2S$