Đáp án:

\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}\,{\rm{HCl}}}} = 35,04g\\
b)\\
{C_\% }FeC{l_2} = 4,95\% \\
{C_\% }FeC{l_3} = 12,66\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
hh:FeO,F{e_2}{O_3}\\
{n_{FeO}} = {n_{F{e_2}{O_3}}} = a\,mol\\
 \Rightarrow 72a + 160a = 3,48 \Rightarrow a = 0,015\,mol\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{HCl}} = 0,015 \times 2 + 0,015 \times 6 = 0,12\,mol\\
{m_{{\rm{dd}}\,{\rm{HCl}}}} = \dfrac{{0,12 \times 36,5}}{{12,5\% }} = 35,04g\\
b)\\
{m_{{\rm{dd}}spu}} = 3,48 + 35,04 = 38,52g\\
{n_{FeC{l_2}}} = {n_{FeO}} = 0,015\,mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,03\,mol\\
{C_\% }FeC{l_2} = \dfrac{{0,015 \times 127}}{{38,52}} \times 100\%  = 4,95\% \\
{C_\% }FeC{l_3} = \dfrac{{0,03 \times 162,5}}{{38,52}} \times 100\%  = 12,66\% 
\end{array}\)

Đáp án:

$a,\ m_{dd\ HCl}=35,04(g)\\ b,\\ C\%_{dd\ FeCl_2}=4,95\%\\ C\%_{dd\ FeCl_3}=12,66\%$

Giải thích các bước giải:

$a,$

 Do $n_{FeO}=n_{Fe_2O_3}$ nên ta quy hỗn hợp thành $Fe_3O_4$

$n_{Fe_3O_4}=\dfrac{3,48}{232}=0,015(mol)\\ Fe_3O_4+8HCl\to FeCl_2+2FeCl_3+4H_2O\\ n_{HCl}=8.n_{Fe_3O_4}=0,12(mol)\\ m_{dd\ HCl}=\dfrac{0,12.36,5}{12,5\%}=35,04(g)$

$b,\\ n_{FeCl_2}=n_{Fe_3O_4}=0,015(mol)\\ m_{FeCl_2}=0,015.127=1,905(g)\\ n_{FeCl_3}=2.n_{Fe_3O_4}=0,03(mol)\\ m_{FeCl_3}=0,03.162,5=4,875(g)\\ m_{dd\ sau}=m_{Fe_3O_4}+m_{dd\ HCl}\\ =3,48+35,04\\ =38,52(g)\\ C\%_{dd\ FeCl_2}=\dfrac{1,905}{38,52}.100\%=4,95\%\\ C\%_{dd\ FeCl_3}=\dfrac{4,875}{38,52}.100\%=12,66\%$