Giải thích các bước giải:

\(\begin{array}{l}
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = 0,3mol\\
{n_{Cu}} = 0,1mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
 \to {m_{Al}} = 5,4g\\
 \to {m_{CuO}} = 19,8 - 5,4 - 6,4 = 8g \to {n_{CuO}} = 0,1mol\\
a)\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} + \dfrac{3}{2}{n_{Al}} = 0,4mol\\
 \to {m_{{H_2}S{O_4}}} = 39,2g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{39,2}}{{17,5\% }} \times 100\%  = 224g\\
b)\\
\% {m_{Cu}} = \dfrac{{6,4}}{{19,8}} \times 100\%  = 32,32\% \\
\% {m_{CuO}} = \dfrac{8}{{19,8}} \times 100\%  = 40,40\% \\
\% {m_{Al}} = \dfrac{{5,4}}{{19,8}} \times 100\%  = 27,27\% \\
c)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + 2{H_2}O + S{O_2}\\
{n_{{H_2}S{O_4}}} = 0,3mol\\
{n_{Cu}} = 0,1mol\\
 \to {n_{Cu}} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{S{O_2}}} = {n_{Cu}} = 0,1mol\\
 \to {V_{S{O_2}}} = 2,24l
\end{array}\)