Giải thích các bước giải:

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
 \to {m_{Al}} = 0,2 \times 27 = 5,4g\\
 \to {m_{Cu}} = 11,8 - 5,4 = 6,4g \to {n_{Cu}} = 0,1mol\\
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + 2{H_2}O + S{O_2}\\
{n_{S{O_2}}} = {n_{Cu}} = 0,1mol\\
 \to {V_{S{O_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} = 0,2mol\\
 \to {m_{{H_2}S{O_4}}} = 0,2 \times 98 = 19,6g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = (\dfrac{{19,6}}{{98\% }} \times 100\% ) + 10\%  = 20,1g\\
b)\\
{m_{{\rm{dd}}NaOH}} = 25 \times 1,28 = 32g\\
 \to {m_{NaOH}} = \dfrac{{32 \times 25\% }}{{100\% }} = 8g\\
 \to {n_{NaOH}} = 0,2mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,2}}{{0,1}} = 2
\end{array}\)

-> Tạo 1 muối trung hòa: Na2SO3

\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,1mol\\
 \to {m_{N{a_2}S{O_3}}} = 0,1 \times 126 = 12,6g\\
 \to {m_{{\rm{dd}}N{a_2}S{O_3}}} = {m_{S{O_2}}} + {m_{{\rm{dd}}NaOH}} = 0,1 \times 64 + 32 = 38,4g\\
 \to C{\% _{{\rm{dd}}N{a_2}S{O_3}}} = \dfrac{{12,6}}{{38,4}} \times 100\%  = 32,8\% \\

\end{array}\)