Tổng quát:
\(OH^{^-}+H^{^+}->H_2O\\ Ba^{^{2+}}+SO_4^{^{2-}}->BaSO_4\downarrow\\ n_{OH^-}=2.1,2.0,2=0,48mol=n_{H^+}\\ m_{ddX}=a\left(g\right)\\ n_{H^+}=0,48=\dfrac{0,146a}{36,5}+\dfrac{0,196a}{98}.2\\ a=60\left(g\right)\\ m_{BaSO_4}=233.1,2.0,2=55,92g\)