Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{CuO}} = 61,07\% \\
\% {m_{A{l_2}{O_3}}} = 38,93\% \\
b)\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 34,2g\\
{m_{CuS{O_4}}} = 32g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
hh:A{l_2}{O_3}(a\,mol),CuO(b\,mol)\\
{n_{{H_2}S{O_4}}} = 0,25 \times 2 = 0,5\,mol\\
\left\{ \begin{array}{l}
102a + 80b = 26,2\\
3a + b = 0,5
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,2\\
\% {m_{CuO}} = \dfrac{{0,2 \times 80}}{{26,2}} \times 100\%  = 61,07\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 61,07 = 38,93\% \\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,1\,mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,2\,mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g\\
{m_{CuS{O_4}}} = 0,2 \times 160 = 32g
\end{array}\)