Em tham khảo nha :

\(\begin{array}{l}
4)\\
a)\\
{n_{KOH}} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{C_{{M_{KOH}}}}\dfrac{{0,1}}{{0,4}} = 0,25M\\
b)\\
{C_{{M_{HN{O_3}}}}} = \dfrac{{0,5}}{{0,2}} = 2,5M\\
c)\\
{C_{{M_{NaOH}}}} = \dfrac{{0,25}}{{0,25}} = 1M\\
d)\\
{n_{{H_2}S{O_4}}} = \dfrac{{14,7}}{{98}} = 0,15mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,15}}{{0,18}} = \frac{5}{6}M\\
e)\\
{n_{NaN{O_3}}} = \dfrac{{38,25}}{{85}} = 0,45mol\\
{C_{{M_{NaN{O_3}}}}} = \dfrac{{0,45}}{{0,27}} = \frac{5}{3}M\\
5)\\
a)\\
{n_{NaOH}} = \dfrac{4}{{40}} = 0,1mol\\
{V_{NaOH}} = \dfrac{{0,1}}{2} = 0,05l = 50ml\\
b)\\
{V_{FeC{l_3}}} = \dfrac{2}{{0,15}} = \frac{{40}}{3}l\\
c)\\
{V_{MgS{O_4}}} = \dfrac{{0,2}}{{0,8}} = 0,25l = 250ml\\
d)\\
{n_{HCl}} = \dfrac{{7,3}}{{36,5}} = 0,2mol\\
{V_{HCl}} = \dfrac{{0,2}}{{0,4}} = 0,5l = 500ml\\
6)\\
a)\\
{n_{Ba{{(OH)}_2}}} = 0,25 \times 2 = 0,5mol\\
b)\\
{n_{FeC{l_3}}} = 0,08 \times 0,15 = 0,012mol\\
c)\\
{n_{MgS{O_4}}} = 4,5 \times 0,8 = 3,6mol\\
d)\\
{n_{Zn{{(N{O_3})}_2}}} = 0,015 \times 0,4 = 0,006mol
\end{array}\)