$n_{Mg}=4,5/24=0,1875mol$
$a.Mg+2HCl\to MgCl_2+H_2↑(1)$
b.Theo pt (1) :
$n_{H_2}=n_{Mg}=0,1875mol$
$⇒V_{H_2}=0,1875.22,4=4,2l$
$c.2Na+2H_2O\to 2NaOH+H_2↑(2)$
Theo pt (2) :
$n_{Na}=2.n_{H_2}=2.0,1875=0,375mol$
$⇒m_{Na}=0,375.23=8,625g$
d.Theo pt (1) :
$n_{HCl}=2.n_{Mg}=2.0,1875=0,375mol$
$⇒m_{HCl}=0,375.36,5=13,6875g$
Vì lượng axit lấy dư 10%
$⇒m_{HCl\ tt}=13,6875+13,6875.10\%=15,05625g$
$⇒m_{dd\ HCl}=\dfrac{15,05625}{7,3\%}=206,25g$