Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=36\left(g\right)\)

\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)

PTHH: 

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,2

Vậy :

\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)

Gọi x, y lần lượt là sô mol của Fe và Mg

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 

Fe + H₂SO₄ ---> FeSO₄ + H₂ (1)

Mg + H₂SO₄ ---> MgSO₄ + H₂ (2)

a. Theo PT₍₁₎: \(n_{H_2}=n_{Fe}=x\left(mol\right)\)

Theo PT₍₂₎: \(n_{H_2}=n_{Mg}=y\left(mol\right)\)

\(\Rightarrow x+y=0,3\) (*)

Theo đề, ta có: 56x + 24y = 10.4 (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}x+y=0,3\\56x+24y=10,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{Mg}=0,2.24=4,8\left(g\right)\)

b. Ta có: \(n_{hh}=0,1+0,2=0,3\left(mol\right)\)

Theo PT_(1,2): \(n_{H_2SO_4}=n_{hh}=0,3\left(mol\right)\)

Đổi 200ml = 0,2 lít

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=39\left(g\right)\)

\(\Rightarrow m_{Al}+m_{Fe}=39\\ \Rightarrow27x+56y=39\left(1\right)\)

\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....x\rightarrow..0.75x....0,5x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow..\dfrac{2}{3}y.....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)

\(\Rightarrow0,75x+\dfrac{2}{3}y=0,55\left(2\right)\)

\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}27x+56y=39\\0,75x+\dfrac{2}{3}y=0,55\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,6.56=33,6\left(g\right)\end{matrix}\right.\\ m_r=m_{Al_2O_3}+m_{Fe_3O_4}=0,5.0,2.102+\dfrac{1}{3}.0,6.232=56,6\left(g\right)\)

Ta có nH2 = 3,36/22,4 = 0,15 mol
        Fe +2 HCl -> FeCl2 + H2
       0,15.    0,3                   <-. 0,15.  ( Mol)
=> mFe = 0,15 × 56 = 8,4g
   => %Fe = 8,4/15×100% = 56% 
      => %Cu = 100% - 56% = 44%

=>VHCl =1\0,3=10\3 l

PTHH :  2Fe    +     6HCl   -->   2FeCl₃  +  3H₂   (1)

n_H2 =  \(\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

Từ (1) ->  n_Fe = \(\dfrac{2}{3}n_{H_2}=0.1\left(mol\right)\)

-> m_Fe = n.M = 0,1  . 56 = 5.6 (g) => %m_Fe = \(\dfrac{5.6}{15}x100\%\approx37.3\%\)

-> %m_Cu =  100% - 37.3% = 62.7 %