\(n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\)
\(Ba\left(OH\right)_2+Na_2SO_4\rightarrow BaSO_4+2NaOH\)
0,1 -----------> 0,1------------>0,1----------> 0,2
\(m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)
C1: \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
c2: theo đlbtkl \(m_{Na_2SO_4}=23,3+8-17,1=14,2\left(g\right)\)