$P_{(Coloured \ Solution)} + BaCl_2 \to Q \downarrow_{(White)} + R_{(Coloured \ Solution)}$
Then salt $P$ in above reaction is

Match List-$I$ with List-$II$:

List-$I$ (Anion)	List-$II$ (Gas evolved on reaction with dil. $H_2SO_4$)
$A. CO_3^{2-}$	$I$. Colourless gas which turns lead acetate paper black
$B. S^{2-}$	$II$. Colourless gas which turns acidified potassium dichromate solution green
$C. SO_3^{2-}$	$III$. Brown fumes which turn acidified $KI$ solution containing starch blue
$D. NO_2^{-}$	$IV$. Colourless gas evolved with brisk effervescence,which turns lime water milky

Choose the correct answer from the options given below:

$A$ white precipitate was formed when $BaCl_{2}$ was added to the water extract of an inorganic salt. Further,a gas $X$ with a characteristic odour was released when the formed white precipitate was dissolved in dilute $HCl$. The anion present in the inorganic salt is ..... .

The reagent$(s)$ that can selectively precipitate $S^{2-}$ from a mixture of $S^{2-}$ and $SO_4^{2-}$ in aqueous solution is(are):
$(A)$ $CuCl_2$
$(B)$ $BaCl_2$
$(C)$ $Pb(CH_3COO)_2$
$(D)$ $Na_2[Fe(CN)_5NO]$

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

When $BaCl_2$ is added to a clear solution of a compound $(X)$,a heavy white precipitate is formed which does not dissolve in $dil. HCl$. What is the compound $(X)$?