Em tham khảo nha:
\(\begin{array}{l}
4)\\
{n_{KOH}} = 0,5 \times 1 = 0,5\,mol\\
{n_{HCl}} = {n_{KOH}} = 0,5\,mol\\
{m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{18,25}}{{7,3\% }} = 250g\\
5)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{m_{NaOH}} = 20 \times 10\% = 2g\\
{n_{NaOH}} = \dfrac{2}{{40}} = 0,05\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,05}}{2} = 0,025\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,025}}{{0,04}} = 0,625M
\end{array}\)
