$n_{Mg}=4,8/24=0,2mol$
$a.Mg+2HCl\to MgCl_2+H_2↑(1)$
b.Theo pt (1) :
$n_{H_2}=n_{Mg}=0,2mol$
$⇒V_{H_2}=0,2.22,4=4,48l$
$c.2Na+2H_2O\to 2NaOH+H_2↑(2)$
Theo pt (2) :
$n_{Na}=2.n_{H_2}=2.0,2=0,4mol$
$⇒m_{Na}=0,4.23=9,2g$
d.Theo pt (1) :
$n_{HCl}=2.n_{Mg}=2.0,2=0,4mol$
$⇒m_{HCl}=0,4.36,5=14,6g$
Vì lượng axit lấy dư 10%
$⇒m_{HCl\ tt}=14,6+14,6.10\%=16,06g$
$⇒m_{dd\ HCl}=\dfrac{16,06}{7,3\%}=220g$
