Question

A ______ (black/white) precipitate of ______ ___(barium sulphate/ barium bisulphate) is formed when barium chloride is treated with sulphuric acid.

Answer 1

1. When barium chloride (${\mathrm{BaCl}}_2$) is treated with sulphuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$), a white precipitate of barium sulphate (${\mathrm{BaSO}}_4$) along with hydrochloric acid ($\mathrm{HCl}$) is formed.
2. $$\begin{gathered} {\mathrm{BaCl}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{BaSO}}_4\left(\mathrm{aq}\right)+2\mathrm{HCl}\left(\mathrm{l}\right) \\ \mathrm{Barium}\mathrm{chloride}+\mathrm{sulphuric}\mathrm{acid}\to \mathrm{Barium}\mathrm{sulphate}+\mathrm{hydrochloric}\mathrm{acid} \end{gathered}$$

So, A white precipitate of barium sulphate is formed when barium chloride is treated with sulphuric acid.