Em tham khảo nha:
\(\begin{array}{l}
1)\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O\\
2NaOH + {N_2}{O_5} \to 2NaN{O_3} + {H_2}O\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
2)\\
{n_{HCl}} = \dfrac{{300 \times 7,3\% }}{{36,5}} = 0,6\,mol\\
{n_{NaOH}} = \dfrac{{200 \times 4\% }}{{40}} = 0,2\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} < {n_{HCl}} \Rightarrow HCl \text{ dư }\\
{n_{HCl}} \text{ dư } = 0,6 - 0,2 = 0,4\,mol\\
{n_{NaCl}} = {n_{NaOH}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 300 + 200 = 500g\\
{C_\% }HCl = \dfrac{{0,4 \times 36,5}}{{500}} \times 100\% = 2,92\% \\
{C_\% }NaCl = \dfrac{{0,2 \times 58,5}}{{500}} \times 100\% = 2,34\%
\end{array}\)
