Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{CuO}} = 70,18\% \\
\% {m_{A{l_2}{O_3}}} = 29,82\% \\
b)\\
{C_M}CuC{l_2} = 0,33M\\
{C_M}AlC{l_3} = 0,22M\\
{C_M}HCl = 0,67M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{HCl}} \text{ ban đầu}= 0,15 \times 2 = 0,3\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HCl}}\text{ dư} = {n_{NaOH}} = 0,1 \times 1 = 0,1\,mol\\
hh:CuO(a\,mol),A{l_2}{O_3}(b\,mol)\\
80a + 102b = 5,7\\
2a + 6b = 0,2\\
 \Rightarrow a = 0,05;b = \dfrac{1}{{60}}\\
\% {m_{CuO}} = \dfrac{{0,05 \times 80}}{{5,7}} \times 100\%  = 70,18\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 70,18 = 29,82\% \\
b)\\
{n_{CuC{l_2}}} = {n_{CuO}} = 0,05\,mol\\
{n_{AlC{l_3}}} = 2{n_{A{l_2}{O_3}}} = \dfrac{1}{{30}}\,mol\\
{C_M}CuC{l_2} = \dfrac{{0,05}}{{0,15}} = 0,33M\\
{C_M}AlC{l_3} = \dfrac{{\dfrac{1}{{30}}}}{{0,15}} = 0,22M\\
{C_M}HCl = \dfrac{{0,1}}{{0,15}} = 0,67M
\end{array}\)