Em tham khảo nha:

\(\begin{array}{l}
1)\\
{n_{NaOH}} = 0,2 \times 1 = 0,2\,mol\\
{n_{HCl}} = 0,2 \times 2 = 0,4\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{{0,2}}{1} < \dfrac{{0,4}}{2} \Rightarrow \text{ HCl dư}\\
{n_{NaCl}} = {n_{NaOH}} = 0,2\,mol\\
{m_{NaCl}} = 0,2 \times 58,5 = 11,7g\\
{n_{HCl}}\text{ dư} = 0,4 - 0,2 = 0,2\,mol\\
{C_M}NaCl = {C_M}HCl \text{ dư}= \dfrac{{0,2}}{{0,2 + 0,2}} = 0,5\,M\\
2)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}} = \dfrac{{200 \times 7,3\% }}{{36,5}} = 0,4\,mol\\
{n_{NaCl}} = {n_{NaOH}} = {n_{HCl}} = 0,4\,mol\\
{m_{NaCl}} = 0,4 \times 58,5 = 23,4g\\
{C_M}NaOH = \dfrac{{0,4}}{{0,2}} = 2M\\
3)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,2\,mol\\
{C_\% }{H_2}S{O_4} = \dfrac{{0,2 \times 98}}{{100}} \times 100\%  = 19,6\% 
\end{array}\)

\(\begin{array}{l}
{n_{{H_2}}} = {n_{Fe}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)