\(n_{BaCl_2}=\frac{400.5,2\%}{208}=0,1\left(mol\right);n_{H_2SO_4}=\frac{100.1,14.19,6\%}{98}=0,228\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
Theo đề: 0,1.........0,228.....................................(mol)
Lập tỉ lệ: \(\frac{0,1}{1}< \frac{0,228}{1}\)=> Sau phản ứng H2SO4 dư
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
dd sau khi lọc bỏ kết tủa: H2SO4 dư, HCl
\(m_{ddsaup.ứ}=400+114-23,3=490,7\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\frac{\left(0,228-0,1\right).98}{490,7}.100=2,56\%\)
\(C\%_{HCl}=\frac{0,1.2.36,5}{490,7}.100=1,49\%\)