Giải thích các bước giải:
\(\begin{array}{l}
4.\\
Ba{(OH)_2} + N{a_2}S{O_4} \to 2NaOH + BaS{O_4}\\
{n_{N{a_2}S{O_4}}} = 0,04mol\\
\to {n_{Ba{{(OH)}_2}}} = {n_{N{a_2}S{O_4}}} = 0,04mol\\
\to {V_{Ba{{(OH)}_2}}} = \frac{{0,04}}{{0,08}} = 0,5l\\
{n_{NaOH}} = 2{n_{N{a_2}S{O_4}}} = 0,08mol\\
\to C{M_{NaOH}} = \dfrac{{0,08}}{{0,5 + 0,4}} = 0,089M\\
5.\\
S{O_3} + 2NaOH \to N{a_2}S{O_4} + {H_2}O\\
{n_{S{O_3}}} = 0,08mol\\
\to {n_{NaOH}} = 2{n_{S{O_3}}} = 0,16mol\\
\to {V_{NaOH}} = \dfrac{{0,16}}{{0,64}} = 0,25l\\
\to {n_{N{a_2}S{O_4}}} = {n_{S{O_3}}} = 0,08mol\\
\to C{M_{N{a_2}S{O_4}}} = \dfrac{{0,08}}{{0,25 + 1,792}} = 0,039M\\
6.\\
Al{(OH)_3} + 3HCl \to AlC{l_3} + 3{H_2}O\\
{n_{Al{{(OH)}_3}}} = 0,06mol\\
\to {n_{HCl}} = 3{n_{Al{{(OH)}_3}}} = 018mol\\
\to {m_{HCl}} = 6,57g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{6,57}}{{14,6\% }} \times 100\% = 45g\\
\to {n_{AlC{l_3}}} = {n_{Al{{(OH)}_3}}} = 0,06mol \to {m_{AlC{l_3}}} = 8,01g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{8,01}}{{4,68 + 45}} \times 100\% = 16,12\% \\
7.\\
MgS{O_4} + 2KOH \to Mg{(OH)_2} + {K_2}S{O_4}\\
a)\\
{m_{KOH}} = \dfrac{{240 \times 0,8\% }}{{100\% }} = 1,92g\\
\to {n_{KOH}} = 0,034mol\\
\to {n_{Mg{{(OH)}_2}}} = \dfrac{1}{2}{n_{KOH}} = 0,017mol \to {m_{Mg{{(OH)}_2}}} = 0,986g\\
b)\\
{n_{{K_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,017mol \to {m_{{K_2}S{O_4}}} = 2,958g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}MgS{O_4}}} + {m_{{\rm{dd}}KOH}} - {m_{Mg{{(OH)}_2}}} = 439,014g\\
\to C{\% _{{K_2}S{O_4}}} = \dfrac{{2,958}}{{439,014}} \times 100\% = 0,67\% \\
c)\\
Mg{(OH)_2} \to MgO + {H_2}O\\
{n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,017mol\\
\to {m_{MgO}} = 0,68g
\end{array}\)
