$m_{BaCl_2}=200.5,2\%=10,4g$
$⇒n_{BaCl_2}=\dfrac{10,4}{208}=0,05mol$
$m_{H_2SO_4}=20\%.58,8=11,76g$
$⇒n_{H_2SO_4}=\dfrac{11,76}{98}=0,12mol$
$PTHH :$
$BaCl_2 + H_2SO_4\to BaSO_4↓+2HCl$
Theo pt : 1 mol 1 mol
Theo đbài : 0,05 mol 0,12 mol
Tỉ lệ : $\dfrac{0,05}{1}<\dfrac{0,12}{1}$
⇒Sau phản ứng H2SO4 dư
$Theo\ pt : \\n_{H_2SO_4\ pư}=n_{BaCl_2}=0,05mol \\⇒n_{H_2SO_4\ dư}=0,12-0,05=0,07mol \\⇒m_{H_2SO_4\ dư}=0,07.98=6,86g \\n_{BaSO_4}=n_{HCl}=n_{BaCl_2}=0,05mol \\⇒m_{BaSO_4}=0,05.233=11,65g \\n_{HCl}=2.n_{BaCl_2}=2.0,05=0,1mol \\⇒m_{HCl}=0,1.36,5=3,65g \\m_{dd\ spu}=200+58,8-11,65=247,15g \\⇒C\%_{H_2SO_4\ dư}=\dfrac{6,86}{247,15}.100\%=2,78\% \\C\%_{HCl}=\dfrac{3,65}{247,15}.100\%=1,47\%$
