$n_{Zn}=9,75/65=0,15mol$
$PTHH :$
$Zn+2HCl\to ZnCl_2+H_2$
Theo pt :
$n_{H_2}=n_{Zn}=0,15mol$
$⇒V_{H_2}=0,15.22,4=3,36l$
$n_{Fe_3O_4}=23,2/232=0,1mol$
$PTHH :$
$Fe_3O_4 +4H_2\overset{t^o}\to 3Fe+4H_2O$
Theo pt : 1 mol 4 mol
Theo đbài: 0,1 mol 0,15 mol
⇒Sau pư Fe4O4 dư 0,0625mol
$⇒m_{rắn}=m_{Fe_3O_4\ dư}+m_{Fe}=0,0625.232+0,1125,65=21,8125g$