C là \(BaSO_4\), D là \(HCl\)
\(a,PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\\ \Rightarrow n_{BaSO_4}=0,15\left(mol\right)\\ \Rightarrow m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\\ b,n_{HCl}=2n_{BaCl_2}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ m_{dd_{HCl}}=31,2+100-34,95=96,25\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{96,25}\cdot100\%\approx11,38\%\)