\(a,m_{BaCl_2}=\dfrac{200.20,8\%}{100\%}=41,6g\\ n_{BaCl_2}=\dfrac{41,6}{208}=0,2mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2mol\\ m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6g\\ b,m_{H_2SO_4}=0,2.98=19,6g\\ C_{\%H_2SO_4}=\dfrac{19,6}{200}\cdot100\%=9,8\%\)