$m_{BaCl_2=\frac{5.2\cdot 400}{100}=20.8\Rightarrow n_{BaCl_2=\frac{20.8}{208}=0.1}}$
$m_{ddH_2SO_4}=\frac{11\cdot 100}{20}=55$ (g)
$n_{H_2SO_4}=\frac{11}{98}$ (mol)
PTHH
$BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl$
Vì $\frac{11}{98}>0.1\Rightarrow H_2SO_4dư$
$\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0.1$
$\Rightarrow m_{\downarrow }=m_{BaSO_4}=0.1\cdot 233=23.3\left(g\right)$
Theo đlbtkl
$m_{ddsaupư}=55+400-23.3=431.7\left(g\right)$ (g)
Theo PT =>$n_{HCl}=2n_{BaCl_2}=2\cdot 0.1=0.2$ (mol)
$\Rightarrow C\%_{HCl}=\frac{2\cdot 0.1\cdot 36.5}{431.7}\cdot 100\%\approx 1.7\%$
$C\%_{H_2SO_4dư}=\frac{\left(\frac{11}{98}-0.1\right)98}{431.7}\cdot 100\%\approx 0.28\%$