Em tham khảo nha :
\(\begin{array}{l}
1)\\
2A + 2xHCl \to 2AC{l_x} + x{H_2}\\
{n_{HCl}} = 2mol\\
{n_A} = \dfrac{{{n_{HCl}}}}{x} = \frac{2}{x}mol\\
{M_A} = 18:\dfrac{2}{x} = 9x\,dvC\\
x = 3 \Rightarrow {M_A} = 27dvC\\
A:\text{Nhôm}(Al)\\
2)\\
2B + 2xHCl \to 2BC{l_x} + x{H_2}\\
{n_B} = \dfrac{{4,48}}{{{M_B}}}\,mol\\
{n_{BC{l_x}}} = \dfrac{{10,16}}{{{M_B} + x{M_{Cl}}}}mol\\
{n_B} = {n_{BC{l_x}}}\\
\Rightarrow \dfrac{{4,48}}{{{M_B}}} = \dfrac{{10,16}}{{{M_B} + x{M_{Cl}}}}\\
x = 2 \Rightarrow {M_B} = 56dvC\\
B:\text{Sắt}(Fe)
\end{array}\)