Mình chừa thói quen ăn nói vậy rồi.
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\\ n_{BaSO_4}=\dfrac{46,6}{233}=0,2\left(mol\right)\\ n_{Ba\left(OH\right)_2}=n_{BaSO_4}=0,2\left(mol\right)\\ C\%_{ddBa\left(OH\right)_2}=\dfrac{0,2.171}{300}.100=11,4\%\)