\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
PTHH: BaCl2 + H2SO4 ---> BaSO4↓ + 2HCl
LTL: 0,1 = 0,1 => pư vừa đủ
Theo pthh: \(\left\{{}\begin{matrix}n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\\n_{HCl}=2n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{BaSO_4}=0,1.233=23,3\left(g\right)\\m_{HCl}=0,2.36,5=7,3\left(g\right)\end{matrix}\right.\)
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