Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 100 \times 1,137 = 113,7g\\
{n_{{H_2}S{O_4}}} = \dfrac{{113,7 \times 20\% }}{{98}} \approx 0,232\,mol\\
{n_{BaC{l_2}}} = 0,1 \times 1 = 0,1\,mol\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
\dfrac{{0,1}}{1} < \dfrac{{0,232}}{1} \Rightarrow \text{$H_2SO_4$ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1\,mol\\
{m_{BaS{O_4}}} = 0,1 \times 233 = 23,3g\\
b)\\
{n_{{H_2}S{O_4}}}\text{ dư} = 0,232 - 0,1 = 0,132\,mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2\,mol\\
{V_{{\rm{dd}}spu}} = 0,1 + 0,1 = 0,2l\\
{C_M}{H_2}S{O_4} \text{ dư} = \dfrac{{0,132}}{2} = 0,066M\\
{C_M}HCl = \dfrac{{0,2}}{2} = 0,1M\\
2)\\
{n_{HCl}} = \dfrac{{300 \times 3,7\% }}{{36,5}} \approx 0,3\,mol\\
{n_{NaOH}} = \dfrac{{200 \times 4\% }}{{40}} = 0,2\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{{0,2}}{1} < \dfrac{{0,3}}{1} \Rightarrow \text{ HCl dư}\\
{n_{HCl}} \text{ dư} = 0,3 - 0,2 = 0,1\,mol\\
{n_{NaOH}} = {n_{NaCl}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 300 + 200 = 500g\\
{C_\% }HCl \text{ dư} = \dfrac{{0,1 \times 36,5}}{{500}} \times 100\% = 0,73\% \\
{C_\% }NaOH = \dfrac{{0,2 \times 40}}{{500}} \times 100\% = 1,6\%
\end{array}\)
