Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2FeC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{3,92}}{{22,4}} = 0,175mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 6,95\\
a + \dfrac{3}{2}b = 0,175
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,05\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{Al}} = 6,95 - 5,6 = 1,35g\\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,05mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,35mol\\
{V_{HCl}} = \dfrac{{0,35}}{{0,5}} = 0,7l\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,1}}{{0,7}} = \dfrac{1}{7}M\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,05}}{{0,7}} = \dfrac{1}{{14}}M\\
c)\\
MO + {H_2} \to M + {H_2}\\
{n_{MO}} = {n_{{H_2}}} = 0,175mol\\
{M_{MO}} = \dfrac{{14}}{{0,175}} = 80dvC\\
 \Leftrightarrow {M_M} = 80 - 16 = 64dvC\\
M:\text{Đồng}(Cu)\\
CTHH:CuO
\end{array}\)