\(\begin{array}{l}
Dat\,KL\,chung\,la\,R\\
n_{H_2}=\frac{6,72}{22,4}=0,3(mol)\\
R+2HCl\to RCl_2+H_2\\
Theo\,PT:\,n_R=n_{H_2}=0,3(mol)\\
\to M_R=\frac{8,86}{0,3}\approx 29,53(g/mol)\\
\to 2\,kim\,loai\,la\,Mg,Ca\\
Dat\,n_{Mg}=x(mol);n_{Ca}=y(mol)\\
\to \begin{cases} x+y=0,3\\ 24x+40y=8,86 \end{cases}\to\begin{cases} x=0,19625\\ y=0,10375 \end{cases}\\
\to \begin{cases} \%m_{Mg}=\frac{24x}{8,86}.100\%\approx 53,16\%\\ \%m_{Ca}=100-53,16=46,84\% \end{cases}
\end{array}\)

Gọi hỗn hợp 2 kim loại đó là \(R\)

\(n_{H_2^{}}^{} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)

\(R + 2HCl \to RCl_2^{} + H_2^{}\)

\({\text{ 0}}{\text{,3                          0}}{\text{,3}}\)

\(M_{R_{TrungBinh}^{}}^{} = \frac{{8,86}}{{0,3}} = 29,53g/mol\)

\({\text{Vi hh 2 kim loai nam ke tiep nhau trong bang tuan hoan va co hoa tri 2 nen}}\)

\(R = Mg,Ca\)

\(Mg:x{\text{ mol; Ca : y mol}}\)

\(24x + 40y = 8,86\)

\(x + y = 0,3\)

\( \to x = 0,19625 \to \% m_{Mg}^{} = \frac{{0,19625.24}}{{8,86}}.100 = 53,16\% \)

\( \to \% m_{Ca}^{} = 100 - 53,16 = 46,84\% \)