1/
$n_{Fe}=5,6/56=0,1mol$
$Fe2O3+3H2→2Fe+3H2O$
$\text{Theo pt :}$
$n_{Fe2O3}=1/2.n_{Fe}=1/2.0,1=0,05mol$
$⇒m_{Fe2O3}=0,05.160=8g$
2/
$\text{Đổi 448ml=0,448l}$
$n_{H_2}=0,448/22,4=0,02 mol$
$\text{PTHH:}$
$A+H2SO4→ASO_4+H2$
$\text{Theo pt :}$
$n_A=n_{H_2}=0,02 mol$
$⇒ M_A=\dfrac{1,3}{0,02}=65g/mol$
$\text{Vậy A là Zn}$