Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Cu + 2{H_2}S{O_4} \to C{\rm{uS}}{O_4} + 2{H_2}O + S{O_2}\\
{n_{{H_2}}} = 0,15mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
 \to {m_{Al}} = 0,1 \times 27 = 2,7g\\
{n_{S{O_2}}} = 0,125mol\\
 \to {n_{Cu}} = 0,125mol\\
 \to {m_{Cu}} = 0,125 \times 64 = 8\\
 \to {m_{hỗnhợp}} = 10,7g\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} + 2{n_{Cu}} = 0,4mol\\
 \to {m_{{H_2}S{O_4}}} = 0,4 \times 98 = 39,2g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{39,2}}{{52\% }} \times 100\%  = 75,38g\\
c)\\
{m_{NaOH}} = \dfrac{{16 \times 25\% }}{{100\% }} = 4g\\
 \to {n_{NaOH}} = 0,1mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,125}} = 0,8
\end{array}\)

-> Tạo 1 muối axit: \(NaHS{O_3}\)

\(\begin{array}{l}
S{O_2} + NaOH \to NaHS{O_3}\\
{n_{NaHS{O_3}}} = {n_{S{O_2}}} = 0,125mol\\
 \to {m_{NaHS{O_3}}} = 0,125 \times 104 = 13g
\end{array}\)