$m_{H_2SO_4}=98.20\%=19,6g \\⇒n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol \\m_{BaCl_2}=5,2\%.400=20,8g \\⇒n_{BaCl_2}=\dfrac{20,8}{208}=0,1mol \\a.PTHH :$
$BaCl_2 + H_2SO_4\to BaSO_4↓+2HCl$
Theo pt : 1 mol 1 mol
Theo đbài : 0,1 mol 0,2 mol
Tỉ lệ : $\dfrac{0,1}{1}<\dfrac{0,2}{1}$
⇒Sau phản ứng H2SO4 dư
Theo pt :
$n_{BaSO_4}=n_{BaCl_2}=0,1mol \\⇒m_{BaSO_4}=0,1.233=23,3g \\c.Theo\ pt : \\n_{H_2SO_4\ pư}=n_{BaCl_2}=0,1mol \\⇒n_{H_2SO_4\ dư}=0,2-0,1=0,1mol \\⇒m_{H_2SO_4\ dư}=0,1.98=9,8g \\n_{HCl}=2.n_{BaCl_2}=2.0,1=0,2mol \\⇒m_{HCl}=0,2.36,5=7,3g \\m_{dd\ spư}=98+400=498g \\⇒C\%_{H_2SO_4\ dư}=\dfrac{9,8}{498}.100\%=1,98\% \\C\%_{HCl}=\dfrac{7,3}{498}.100\%=1,47\%$