$n_{HCl}=0,2.3=0,6mol$
$⇒m_{HCl}=0,6.36,5=21,9g$
$Fe_3O_4+8HCl\overset{t^o}\to FeCl_2+4H_2O+2FeCl_3$
a.Theo pt :
$n_{Fe_3O_4}=1/8.n_{HCl}=1/8.0,6=0,075mol$
$⇒m_{Fe_3O_4}=0,075.232=17,4g$
b.Theo pt :
$n_{FeCl_2}=1/8.n_{HCl}=1/8.0,6=0,075mol$
$⇒m_{FeCl_2}=0,075.127=9,525g$
$n_{FeCl_3}=1/4.n_{HCl}=1/4.0,6=0,15mol$
$⇒m_{FeCl_3}=0,1.162,5=16,25g$
$m_{ddspu}=17,4+21,9=36,6g$
$⇒C\%_{FeCl_2}=\dfrac{9,525}{36,6}.100\%=26,02\%$
$C\%_{FeCl_3}=\dfrac{16,25}{36,6}.100\%=44,52\%$