$2Na+H_2SO_4\to Na_2SO_4+H_2$
$Mg+H_2SO_4\to MgSO_4+H_2$
Giả sử thoát ra 4,5 mol hidro.
$m_{H_2}=4,5.2=9g$
$\Rightarrow m_{dd H_2SO_4}=9:4,5\%=200g$
Gọi x là mol $H_2SO_4$
Theo PTHH, x mol $H_2SO_4$ tạo x mol $H_2$
$m_{H_2SO_4}= 98x$
$\Rightarrow m_{H_2O}=200-98x$
$\Rightarrow n_{H_2O}=\dfrac{200-98x}{18} mol$
$2Na+2H_2O\to 2NaOH+H_2$
$\Rightarrow n_{H_2}=\dfrac{200-98x}{36}$
$\Rightarrow \dfrac{200-98x}{36}+x=4,5$
$\Leftrightarrow x=\dfrac{19}{31}$
$C\%_{H_2SO_4}=\dfrac{98x.100}{200}=30\%$