\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)

Ta có :

\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)

Xét P2 : Chỉ có Al phản ứng với NaOH => nAl.3 = 2nH\(_2\) ( Bảo toàn e)

  => nAl = 0,1 mol

Xét P1 : Fe và Al phản ứng với HCl đặc => 2nFe + 3nAl = 2nH\(_2\)

=> nFe = 0,1 mol

=> Trong mỗi phần thì có : mCu = \(\dfrac{1}{2}.10-27.0,1-56.0,1=1,7\left(g\right)\) 

\(\%m_{Cu\left(X\right)}=\%m_{Cu\left(1/2X\right)}=17\%\)

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

Gọi số mol Mg, Fe, Al là a, b, c

=> 24a + 56b + 27c = 23,8

PTHH: Mg + 2HCl --> MgCl₂ + H₂ 

            a------------------------->a

            Fe + 2HCl --> FeCl₂ + H₂

            b------------------------->b

            2Al + 6HCl --> 2AlCl₃ + 3H₂ 

            c------------------------->1,5c

=> a + b + 1,5c = \(\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

PTHH: Mg + Cl₂ --t^o--> MgCl₂

             a-->a

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

             b--->1,5b

             2Al + 3Cl₂ --t^o--> 2AlCl₃

             c--->1,5c

=> \(a+1,5b+1,5c=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

=> a = 0,3; b = 0,2; c = 0,2

=> \(\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

Đáp án C