\(n_{BaSO_4}=\dfrac{58.25}{233}=0.25\left(mol\right)\)
\(BaCl_2+SO_3+H_2O\rightarrow BaSO_4+2HCl\)
\(0.25........0.25.......................0.25........0.5\)
\(V_{SO_3}=0.25\cdot22.4=5.6\left(l\right)\)
\(m_{dd_{BaCl_2}}=\dfrac{0.25\cdot208}{20\%}=260\left(g\right)\)
\(m_{dd}=m_{SO_3}+m_{dd_{BaCl_2}}-m_{BaSO_4}=0.25\cdot80+260-58.25=221.75\left(g\right)\)
\(C\%_{HCl}=\dfrac{0.5\cdot36.5}{221.75}\cdot100\%=8.2\%\)