When BaCl_2 is added to a clear solution of c | vedclass

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(C) The reaction of $BaCl_2$ with a sulphate salt $(SO_4^{2-})$ produces a white precipitate of barium sulphate $(BaSO_4)$.$Ba^{2+}(aq) + SO_4^{2-}(aq

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Sulphate

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(C) The reaction of $BaCl_2$ with a sulphate salt $(SO_4^{2-})$ produces a white precipitate of barium sulphate $(BaSO_4)$.$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) \downarrow$ (White precipitate).$BaSO_4$ is known to be insoluble in dilute $HCl$ and dilute $HNO_3$.Carbonates $(CO_3^{2-})$ also form white precipitates with $BaCl_2$,but they are soluble in dilute $HCl$ with the evolution of $CO_2$ gas.Therefore,the compound $(X)$ must be a sulphate.

When $BaCl_2$ is added to a clear solution of compound $(X)$,a heavy white precipitate is formed,which is insoluble in dilute $HCl$. What is compound $(X)$?

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