\begin{array}{l} BaC{l_2} + N{a_2}S{O_4} \to 2NaCl + B{\rm{aS}}{O_4}\\ {n_{BaC{l_2}}} = 0,1mol\\ {n_{N{a_2}S{O_4}}} = 0,2mol\\ \to {n_{BaC{l_2}}} < {n_{N{a_2}S{O_4}}} \to {n_{N{a_2}S{O_4}}}dư\\ \to {n_{N{a_2}S{O_4}}}(pt) = {n_{BaC{l_2}}} = 0,1mol\\ \to {n_{N{a_2}S{O_4}}}(dư) = 0,1mol\\ {n_{{\rm{BaS}}{O_4}}} = {n_{BaC{l_2}}} = 0,1mol \to {m_{{\rm{BaS}}{O_4}}} = 23,3g\\ {n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\ \to C{M_{NaCl}} = \dfrac{{0,2}}{{0,1 + 0,2}} = 0,67M\\ \to C{M_{N{a_2}S{O_4}}}(dư) = \dfrac{{0,1}}{{0,1 + 0,2}} = 0,33M\\ \end{array}