a) Gọi  X₂O_y  là oxit kim loại cần tìm

Ta có: M_O =30% M_X2Oy 

=> y*16=30% *160=48

=> y=3

 Có: M_X2Oy = 160 (g/mol)

=> 2X +16y=160

=>2X +16*3=160

=>2X =112

=> X=56  => X là Fe

Vậy oxit kim loại cần tìm là Fe₂O₃

b)

n_Fe2O3=8 / 160=0.05(mol)

m_HCl=131.4 * 25%=32.85(g)

=>n_HCl =32.85 /36.5=0.9(mol)

PTHH Fe₂O₃ +6HCl --> 2FeCl₃ +3H₂O

Vì 0.05 < 0.9 / 6=>Fe₂O₃ PƯ hết => dd A gồm FeCl₃ và HCl dư

=> n_HCl PU  =6n_Fe2O3 =6*0.05=0.3(mol)

=> n_{HCl du} =0.9-0.3=0.6(mol)

=> m_{ddHCl du} =0.6*36.5=21.9(g)

Theo PTHH n_FeCl3 =2n_Fe2O3 =2*0.05=0.1(mol)

            => m_FeCl3 0.1* 162.5=16.25(g)

Theo PTHH n_H2O =3n_Fe2O3 =3*0.05=0.15(mol)

           => m_H2O =0.15*18=2.7(g)

Theo ĐLBT KL  Có m_Fe2O3 +m_ddHCl = m_dd A + m_H2O - m_{HCl du}

   => m_dd A =8+131.4 - 2.7 -21.9= 114.8

=> C%_ddHCl =21.9 / 114.8 *100%=19.08%

=> C%_Fe2O3 =16.25 / 114.8 *100% =14.16%

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CTHH:F{e_x}{O_y}\\
y = \dfrac{{160 \times 30\% }}{{16}} = 3\\
x = \dfrac{{160 - 3 \times 16}}{{56}} = 2\\
 \Rightarrow CTHH:F{e_2}{O_3}\\
b)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{m}{M} = \dfrac{8}{{160}} = 0,05mol\\
{m_{HCl}} = \dfrac{{131,4 \times 25\% }}{{100}} = 32,85g\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{32,85}}{{36,5}} = 0,9mol\\
\dfrac{{0,05}}{1} < \dfrac{{0,9}}{6} \Rightarrow HCl\text{ dư}\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
{m_{FeC{l_3}}} = 0,1 \times 162,5 = 16,25g\\
{n_{HC{l_d}}} = {n_{HCl}} - 6{n_{F{e_2}{O_3}}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
{m_{ddspu}} = 8 + 131,4 = 139,4g\\
C{\% _{FeC{l_3}}} = \dfrac{{16,25}}{{139,4}} \times 100\%  = 11,66\% \\
C{\% _{HC{l_d}}} = \dfrac{{21,9}}{{139,4}} \times 100\%  = 15,71\% 
\end{array}\)