Em tham khảo nha :
\(\begin{array}{l}
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{m_{BaC{l_2}}} = \dfrac{{200 \times 20,8}}{{100}} = 41,6g\\
{n_{BaC{l_2}}} = \dfrac{{41,6}}{{208}} = 0,2mol\\
{m_{N{a_2}S{O_4}}} = \dfrac{{300 \times 14,2}}{{100}} = 42,6g\\
{n_{N{a_2}S{O_4}}} = \dfrac{{42,6}}{{142}} = 0,3mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,3}}{1} \Rightarrow N{a_2}S{O_4}\text{ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,2mol\\
{m_{BaS{O_4}}} = 0,2 \times 233 = 46,6g\\
{m_{{\rm{ddspu}}}} = 200 + 300 - 46,6 = 453,4g\\
{n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,4mol\\
{m_{NaCl}} = 0,4 \times 58,5 = 23,4g\\
{n_{N{a_2}S{O_4}d}} = 0,3 - 0,2 = 0,1mol\\
{m_{N{a_2}S{O_4}d}} = 0,1 \times 142 = 14,2g\\
C{\% _{NaCl}} = \dfrac{{23,4}}{{453,4}} \times 100\% = 5,16\% \\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{14,2}}{{453,4}} \times 100\% = 3,13\%
\end{array}\)
