Đáp án:
$a/$
$FeO+2HCl→FeCl+2+H_2$
$Fe_2O_3+6HCl→2FeCl_3+3H_2O$
$Fe_3O_4+8HCl→2FeCl_2+FeCl_3+4H_2O$
$FeO+2H_2SO_4→FeSO_4+2+H_2$
$Fe_2O_3+3HCl→Fe_2(SO_4)_33H_2O$
$Fe_3O_4+4H_2SO_4→FeSO_4+Fe_2(SO_4)_3+4H_2O$
$b/$
Quy đổi về $Fe$ $O$
Đặt $n_{H_2O}=x(mol)$
Bt $H$:
$n_{HCl}=2x(mol)$
BTKL:
$\frac{156,8}{2}+36,5.2x=155,4+18x$
$⇒x=1,4$
$⇒n_{HCl}=2,8(mol)$
BT$H$:
$n_{H_2O}=1,4(mol)$
Bt $O$:
$n_O=1,4(mol)$
$\%O=\frac{1,4.16}{78,4}.100=28,57\%$
$\%Fe=71,48\%$
$c/$
Ddajt $a$ $b$ laf mol $HCl, H_2SO_4$
$n_{H+}=a+2b=28(1)$
BTKL:
$\frac{156,8}{2}+36,5a+98b=167,9+18.1,4$
$⇒a=1,8$
$b=0,5$
$⇒CM_{HCl}=7,2M$
$CM_{H_2SO_4}=2M$
