Đáp án:
$a)$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
Phản ứng xảy ra:
\[Mg+2HCl\to MgCl_2+H_2\]
Theo PTHH: $n_{Mg}=n_{H_2}=0,3(mol)$
$\to m_{Mg}=0,3.24=7,2(g)$
$\to \%m_{Mg}=\dfrac{7,2}{23,2}.100\%=31\%$
$\to \%m_{Cu}=100\%-31\%=69\%$
$n_{MgCl_2}=n_{Mg}=0,3(mol)$
$\to m_{MgCl_2}=0,3.95=28,5(g)$