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(A) To selectively precipitate $S^{2-}$ from a mixture containing $SO_4^{2-}$,we need a reagent that forms an insoluble sulfide but a soluble sulfate.

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$A, C$

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(A) To selectively precipitate $S^{2-}$ from a mixture containing $SO_4^{2-}$,we need a reagent that forms an insoluble sulfide but a soluble sulfate.$1$. $CuCl_2$: $Cu^{2+}$ reacts with $S^{2-}$ to form $CuS$ (black precipitate),while $CuSO_4$ is soluble in water. Thus,$CuCl_2$ works.$2$. $BaCl_2$: $Ba^{2+}$ reacts with $SO_4^{2-}$ to form $BaSO_4$ (white precipitate),which is insoluble. This does not selectively precipitate $S^{2-}$.$3$. $Pb(CH_3COO)_2$: $Pb^{2+}$ reacts with $S^{2-}$ to form $PbS$ (black precipitate),while $PbSO_4$ is generally considered insoluble in water. However,in many analytical contexts,$PbS$ is significantly less soluble than $PbSO_4$,and $Pb(CH_3COO)_2$ is often used to test for sulfides. Given the options,$A$ and $C$ are the correct choices.$4$. $Na_2[Fe(CN)_5NO]$ is a reagent used for the detection of $S^{2-}$ (forming a violet color),not for precipitation.

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

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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)$ $Na_2[Fe(CN)_5NO]$