Gọi 2 kim loại là $\overline{X}$
$\overline{X} + 2HCl \to XCl_2 + H_2$
$n_{H_2}=\frac{0,672}{22,4}=0,03 (mol)$
$n_{\overline{X}}=\frac{0,88}{X} (mol)$
$n_{\overline{X}}=n_{H_2}$
$→\frac{0,88}{\overline{X}}=0,03$
$→0,03\overline{X}=0,88$
$→\overline{X}=29,33$
$→24<\overline{X}<40$
$→$ 2 kim loại lần lượt là $Mg, Ca$
b)
$Mg + 2HCl \to MgCl_2 + H_2\\a \hspace{5cm}a\\Ca + 2HCl \to CaCl_2 + H_2\\b \hspace{5cm}b$
$→\left\{\begin{array}{}24a+40b=0,88\\a+b=0,03\end{array}\right.$
$→\left\{\begin{array}a=0,02\\b=0,01\end{array}\right.$
$→m_{Mg}=0,02×24=0,48 (g)\\→\%Mg=\frac{0,48}{0,88}×100=54,55\%\\→\%Ca=100-54,55=45,45\%$