Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)
a. PTHH: \(Ba\left(OH\right)_2+Na_2SO_4--->BaSO_4\downarrow+2NaOH\)
b. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)
=> \(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)
c. Theo PT: \(n_{Na_2SO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)
=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
=> \(C_{\%_{Na_2SO_4}}=\dfrac{28,4}{240}.100\%=11,83\%\)