\(Đặt:m_{CuO}=m_{Fe_2O_3}=x\left(g\right)\left(x>0\right)\\ \Rightarrow n_{Fe_2O_3}=\dfrac{x}{160}\left(mol\right);n_{CuO}=\dfrac{x}{80}\left(mol\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{CuCl_2}=n_{CuO}=\dfrac{x}{80}\left(mol\right)\\ m_{CuCl_2}=\dfrac{135.x}{80}\left(g\right)\\ n_{FeCl_3}=2.\dfrac{x}{160}=\dfrac{x}{80}\left(mol\right)\\ m_{FeCl_3}=\dfrac{162,5.x}{80}\left(g\right)\\ \dfrac{m_{CuCl_2}}{m_{FeCl_3}}=\dfrac{\dfrac{135x}{80}}{\dfrac{162,5x}{80}}=\dfrac{54}{65}\)

\(\dfrac{m_{CuO}}{m_{Fe_2O_3}}=1\Rightarrow\dfrac{80n_{CuO}}{160.n_{Fe_2O_3}}=1\Rightarrow\dfrac{n_{CuO}}{n_{Fe_2O_3}}=2\)

$CuO + 2HCl \to CuCl_2 + H_2O$
$Fe_2O_3 + 6HCl \to 2FeCl_3  + 3H_2O$

$n_{CuCl_2} = n_{CuO} ; n_{FeCl_3} = 2n_{Fe_2O_3}$

\(\dfrac{m_{CuCl_2}}{m_{FeCl_3}}=\dfrac{135n_{CuCl_2}}{162,5.n_{FeCl_3}}=\dfrac{135}{162,5}.\dfrac{n_{CuCl_2}}{n_{FeCl_3}}\)

\(=\dfrac{135}{162,5}.\dfrac{n_{CuO}}{2n_{Fe_2O_3}}=\dfrac{135}{162,5}.\dfrac{1}{2}.2=0,83\)

Ta có: \(m_{CuO}=m_{Fe_2O_3}\Leftrightarrow n_{CuO}.80=n_{Fe_2O_3}.160\Leftrightarrow\dfrac{n_{CuO}}{n_{Fe_2O_3}}=2\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Ta có: \(n_{CuCl_2}=n_{CuO};n_{FeCl_3}=2n_{Fe_2O_3}\)

  \(\Rightarrow\dfrac{n_{CuCl_2}}{n_{FeCl_3}}=\dfrac{n_{CuO}}{2n_{Fe_2O_3}}=\dfrac{2}{2}=1\)

Gọi số mol Fe₂O₃, CuO là a, b (mol)

n_HCl = 0,3.2 = 0,6 (mol)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

                 a----->6a------->2a

            CuO + 2HCl --> CuCl₂ + H₂O

              b----->2b------->b

=> \(\left\{{}\begin{matrix}\dfrac{2a}{b}=\dfrac{3}{4}\\6a+2b=0,6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=\dfrac{9}{170}\left(mol\right)\\b=\dfrac{12}{85}\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{\dfrac{9}{170}.160}{\dfrac{9}{170}.160+\dfrac{12}{85}.80}.100\%=42,857\%\\\%m_{CuO}=\dfrac{\dfrac{12}{85}.80}{\dfrac{9}{170}.160+\dfrac{12}{85}.80}.100\%=57,143\%\end{matrix}\right.\)

Đặt \((n_{MgO};n_{CuO})=(a;b)\)

\(MgO+H_2SO_4\to MgSO_4+H_2O\\ CuO+H_2SO_4\to CuSO_4+H_2O\\ \Rightarrow n_{MgSO_4}:n_{CuSO_4}=1:1=a:b\\ \Rightarrow a=b\)

Mà \(a+b=n_{H_2SO_4}=1.0,3=0,3(mol)\)

\(\Rightarrow a=b=0,15(mol)\\ \Rightarrow m_{MgO}=0,15.40=6(g)\)

Chọn B.

Đáp án B

CuO + 2HCl → CuCl₂ + H₂O

Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Giả sử số mol 2 muối CuCl₂ và FeCl₃ đều là 1 mol

=> n_CuO = n_CuCl2 = 1 mol ; n_Fe2O3 = 0,5n_FeCl3 = 0,5 mol

=> m_{hh đầu} = 80.1 + 160.0,5 = 160g

=> %m_CuO = %m_Fe2O3 = 50%

Đáp án B

Giả sử nCuCl₂ = nFeCl₃ = 1

 Chọn B.

Đáp án B

Giả sử nCuCl₂ = nFeCl₃ = 1

=>