1/
$m_{NaOH}=6,2+100.8\%= 14,2g$
$m_{dd}=6,2+100=106,2g$
$\Rightarrow C\%=\frac{14,2.100}{106,2}=13,37\%$
2/
$2A+2nHCl\to 2ACl_n + nH_2$
Theo PTHH, $n_A= n_{ACl_n}$
$\Rightarrow \frac{4,8}{A}=\frac{19}{A+35,5n}$
$\Rightarrow 19A=4,8(A+35,5n)$
$\Leftrightarrow A=12n$
$n=2\Rightarrow A=24(Mg)$
Vậy kim loại là magie.
3/
a, $Fe_xO_y+ yH_2\to xFe+yH_2O$
b,
$n_{Fe}=\frac{3,1}{56}=0,06 mol$
$\Rightarrow n_{Fe_xO_y}=\frac{0,06}{x}$
$M_{Fe_xO_y}=\frac{4,5x}{0,06}=75x=56x+16y$
$\Leftrightarrow 19x=16y$
$\Leftrightarrow \frac{x}{y}=\frac{3}{4}$
Vậy oxit là $Fe_3O_4$