\(2NaOH+H_2SO_4->Na_2SO_4+2H_2O\\ Ba\left(OH\right)_2+H_2SO_4->BaSO_4+2H_2O\\ n_{H^+}=150\cdot\dfrac{0,196}{98}\cdot2=0,6mol=m\left(\dfrac{0,048}{40}+2\cdot\dfrac{0,1539}{171}\right)\\ m=200\left(g\right)\\ n_{Ba^{2+}}=\dfrac{0,1539}{171}m=0,18mol\\ n_{SO_4^{2-}}=150\cdot\dfrac{0,196}{98}=0,3mol\\ m_1=0,18\cdot233=41,94g\\ C\%_{Na_2SO_4}=\dfrac{0,12\cdot142}{350-41,94}\cdot100\%=5,53\%\)