$n_{HCl}=\dfrac{150.7,3\%}{36,5}=0,3mol$
$PTHH :$
$Mg+2HCl\to MgCl_2+H_2↑$
a/Theo pt :
$n_{Mg}=1/2.n_{HCl}=1/2.0,3=0,15mol$
$⇒m_{Mg}=0,15.24=3,6g$
b/Theo pt :
$n_{H_2}=1/2.n_{HCl}=1/2.0,3=0,15mol$
$⇒V_{H_2}=0,15.22,4=3,36l$
$c/m_{dd spư}=150+3,6-0,15.2=15,3g$
$n_{MgCl_2}=1/2.n_{HCl}=1/2.0,3=0,15mol$
$⇒C\%MgCl_2=\dfrac{0,15.95}{153,3}.100\%=9,3\%$