Em tham khảo nha:
\(\begin{array}{l}
1)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,5 \times 0,12 = 0,06\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,06}}{2} = 0,03\,mol\\
a = {C_M}{H_2}S{O_4} = \dfrac{{0,03}}{{0,8}} = 0,0375M\\
2)\\
Ba{(OH)_2} + 2HN{O_3} \to Ba{(N{O_3})_2} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,6 \times 0,18 = 0,108\,mol\\
{n_{HN{O_3}}} = 0,108 \times 2 = 0,216\,mol\\
V = \dfrac{{0,216}}{{0,15}} = 1,44l\\
3)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,6 \times 0,15 = 0,09\,mol\\
{n_{NaOH}} = 0,09 \times 2 = 0,18\,mol\\
X = {C_\% }NaOH = \dfrac{{0,18 \times 40}}{{200}} \times 100\% = 3,6\% \\
4)\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{Ca{{(OH)}_2}}} = \dfrac{{200 \times 29,6\% }}{{74}} = 0,8\,mol\\
{n_{HCl}} = 2{n_{Ca{{(OH)}_2}}} = 1,6\,mol\\
b = {C_M}HCl = \dfrac{{1,6}}{{0,8}} = 2M
\end{array}\)