a/ $Mg+2HCl\to MgCl_2+H_2$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
Theo phương trình: $n_{Mg}=n_{H_2}$
$→n_{Mg}=0,3(mol)\\→m_{Mg}=0,3.24=7,2g$
Vậy $m_{Mg}=7,2g$
b/ Theo phương trình: $n_{HCl}=2n_{H_2}$
$→n_{HCl}=0,6(mol)\\→m_{HCl}=0,6.36,5=21,9g\\m_{dd\,HCl}=\dfrac{21,9}{10\%}=219g$
Vậy $m_{dd\,HCl}=219g$