Giải thích các bước giải:

7,

Gọi a và b là số mol của Mg và Fe

\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,3mol\\
\left\{ \begin{array}{l}
24a + 56b = 10,4\\
a + b = 0,3
\end{array} \right.\\
a = 0,2\\
b = 0,1\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = {n_{Mg}} + 1,5{n_{Fe}} = 0,35mol \to {V_{S{O_2}}} = 7,84l
\end{array}\)

8,

\(\begin{array}{*{20}{l}}
{Zn + 2HCl \to ZnC{l_2} + {H_2}}\\
{FeO + 2HCl \to FeC{l_2} + {H_2}O}\\
{{n_{{H_2}}} = 0,2mol}\\
{{n_{Zn}} = {n_{{H_2}}} = 0,2mol \to {m_{Zn}} = 13g}\\
\begin{array}{l}
{m_{FeO}} = 26 - 13 = 13g \to {n_{FeO}} = 0,18mol\\
\% {m_{Zn}} = \% {m_{FeO}} = 50\% 
\end{array}\\
{{m_{muối}} = {m_{ZnC{l_2}}} + {m_{FeC{l_2}}} = 0,2 \times 136 + 0,18 \times 127 = 50,06g}\\
{{n_{HCl}} = 2{n_{Zn}} + 2{n_{FeO}} = 0,76mol \to {V_{HCl}} = \dfrac{{0,76}}{2} = 0,38l}
\end{array}\)

9,

a,

\(\begin{array}{l}
{n_{{H_2}S}} = 0,15mol\\
{n_{KOH}} = 0,4mol\\
\dfrac{{{n_{KOH}}}}{{{n_{{H_2}S}}}} = 2,67
\end{array}\)

Tạo ra muối trung hòa, KOH dư

\(\begin{array}{l}
2KOH + {H_2}S \to {K_2}S + 2{H_2}O\\
{n_{{K_2}S}} = {n_{{H_2}S}} = 0,15mol\\
{n_{KOH}}dư= 0,4 - 2 \times 0,15 = 0,1mol\\
C{M_{{K_2}S}} = \dfrac{{0,15}}{{0,4}} = 0,375M\\
C{M_{KOH}}dư = \dfrac{{0,1}}{{0,4}} = 0,25M
\end{array}\)

b,

\(\begin{array}{l}
{n_{S{O_2}}} = 0,4mol\\
{n_{NaOH}} = \dfrac{{200 \times 18}}{{100 \times 40}} = 0,9mol\\
\dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = 2,25
\end{array}\)

tạo ra muối trung hòa, NaOH dư

\(\begin{array}{l}
2NaOH + S{O_2} \to N{a_2}S{O_3} + 2{H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,4mol\\
{n_{NaOH}}dư= 0,9 - 2 \times 0,4 = 0,1mol\\
C{\% _{N{a_2}S{O_3}}} = \dfrac{{0,4 \times 126}}{{0,4 \times 64 + 200}} \times 100\%  = 22,34\% \\
C{\% _{NaOH}}du = \dfrac{{0,1 \times 40}}{{0,4 \times 64 + 200}} \times 100\%  = 1,77\% 
\end{array}\)