Ta có: \(\left\{{}\begin{matrix}n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{147.20\%}{98}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
bđ 0,1 0,3
pư 0,1------->0,1
sau pư 0 0,2 0,1
`=>` \(m_{ddspư}=400+147-0,1.233=523,7\left(g\right)\)
`=>` Chọn B