`n_{BaCl_2}={20,8}/{208}=0,1(mol)`
`n_{H_2SO_4}={20.19,6\%}/{98}=0,04(mol)`
`H_2SO_4+BaCl_2->BaSO_4+2HCl`
`0,04->0,04->0,04->0,08(mol)`
Do `0,1>0,04->BaCl_2` dư.
`C\%_{BaCl_2\ du}={208(0,1-0,04)}/{20,8+20-0,04.233}.100\%\approx 39,64\%`
`C\%_{HCl}={0,08.36,5}/{20,8+20-0,04.233}.100\%\approx 9,28\%`