Gọi x,y,z lần lượt là số mol của Zn, Fe, Cu trong hh đầu

.......k là tỉ lệ của hh X sau so với hh đầu

n_H2 = \(\dfrac{2,24}{22,4}=0,1\) mol

Pt: Zn + 2HCl --> ZnCl₂ + H₂

.......x....................................x

......Fe + 2HCl --> FeCl₂ + H₂

.......y....................................y

Ta có: 65x + 56y + 64z = 9,25 (1)

...........x + y = 0,1 (2)

n_Cl2 = \(\dfrac{7,84}{22,4}=0,35\) mol

Pt: Zn + Cl₂ --to--> ZnCl₂

......kx......kx

.....2Fe + 3Cl₂ --to--> 2FeCl₃

......ky.....1,5ky

......Cu + Cl₂ --to--> CuCl₂

........kz.....kz

Ta có: (x + y + z).k = 0,3

...........(x + 1,5y + z).k = 0,35

\(\Rightarrow\dfrac{\left(x+1,5y+z\right).k}{\left(x+y+z\right).k}=\dfrac{0,35}{0,3}\)

\(\Leftrightarrow\dfrac{x+1,5y+z}{x+y+z}=\dfrac{7}{6}\)

=> x - 2y + z = 0 (3)

Từ (1), (2) và (3) ta có hệ: \(\left\{{}\begin{matrix}65x+56y+64z=9,25\\x+y=0,1\\x-2y+z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z=0,05\)

m_Zn = 0,05 . 65 = 3,25 (g)

m_Fe = 0,05 . 56 = 2,8 (g)

m_Cu = 0,05 . 64 = 3,2 (g)

Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$

Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$

Mặt khác : 

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$

Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$

Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$

Từ (1)(2)(3) suy ra : a = 0,2 ; b = 0,1 ; c = 0,1

$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$

$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$

TN1: Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\end{matrix}\right.\)

=> 65a + 56b + 64c = 37 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

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

            Fe + 2HCl --> FeCl₂ + H₂

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

=> \(a+b=\dfrac{8,96}{22,4}=0,4\) (2)

TN2: Gọi \(\left\{{}\begin{matrix}n_{Zn}=ak\left(mol\right)\\n_{Fe}=bk\left(mol\right)\\n_{Cu}=ck\left(mol\right)\end{matrix}\right.\)

=> ak + bk + ck = 0,15 (3)

\(n_{Cl_2}=\dfrac{3,92}{22,4}=0,175\)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

           ak-->ak

           2Fe + 3Cl₂ --t^o--> 2FeCl₃

            bk--->1,5bk

            Cu + Cl₂ --t^o--> CuCl₂

             ck-->ck

=> ak + 1,5bk + ck = 0,175 (4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\\c=0,2\left(mol\right)\\k=0,25\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{37}.100\%=35,135\%\\\%m_{Fe}=\dfrac{0,2.56}{37}.100\%=30,27\%\\\%m_{Cu}=\dfrac{0,2.64}{37}.100\%=34,595\%\end{matrix}\right.\)

TN1: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(a;b;c\right)\)

=> 24a + 27b + 65c = 28,6 (1)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

             a--->0,5a

            4Al + 3O₂ --t^o--> 2Al₂O₃

             b-->0,75b

            2Zn + O₂ --t^o--> 2ZnO

             c--->0,5c

=> 0,5a + 0,75b + 0,5c = 0,5 (2)

TN2: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(ak;bk;ck\right)\)

=> ak + bk + ck = 0,8 (3)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

           ak----------------------->ak

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

           bk------------------------>1,5bk

           Zn + 2HCl --> ZnCl₂ + H₂

          ck---------------------->ck

=> \(ak+1,5bk+ck=\dfrac{22,4}{22,4}=1\) (4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\\b=0,4\\c=0,2\\k=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{28,6}.100\%=16,783\%\\\%m_{Al}=\dfrac{0,4.27}{28,6}.100\%=37,762\%\\\%m_{Zn}=\dfrac{0,2.65}{28,6}.100\%=45,455\end{matrix}\right.\)

Gọi số mol của Cu, Fe, Al trong 23,8 gam hhX lần lượt là x, y, z mol

→ m_X = 64x + 56y + 27z = 23,8 (1)
nCl2nCl2 = x + 1,5y + 1,5z = 0,65 (2)
0,25 mol X + HCl → 0,2 mol H2 nên 0,2.(x + y + z) = 0,25.(y + 1,5z) (3)
Từ (1), (2), (3) => x = 0,2 mol; y = 0,1 mol; z = 0,2 mol 

%Cu=0,2.6423,8≈53,78%%Cu=0,2.6423,8≈53,78%

%Fe=0,1.5623,8≈23,53%%Fe=0,1.5623,8≈23,53%

 %_Al ≈ 22,69%

Gọi số mol của Cu, Fe, Al trong 23,8 gam hhX lần lượt là x, y, z mol

→ m_X = 64x + 56y + 27z = 23,8 (1)
nCl2nCl2 = x + 1,5y + 1,5z = 0,65 (2)
0,25 mol X + HCl → 0,2 mol H2 nên 0,2.(x + y + z) = 0,25.(y + 1,5z) (3)
Từ (1), (2), (3) => x = 0,2 mol; y = 0,1 mol; z = 0,2 mol 

%Cu=0,2.6423,8≈53,78%%Cu=0,2.6423,8≈53,78%

%Fe=0,1.5623,8≈23,53%%Fe=0,1.5623,8≈23,53%

 %_Al ≈ 22,69%

a)

\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂ (1)

             0,6<----------------------0,3

=> m_Na = 0,6.23 = 13,8 (g)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1<-0,2

=> m_Fe = 0,1.56 = 5,6 (g)

m_Cu = 10 (g)

\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

             \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)

=> 56x = 42y

=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe₃O₄

a)

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

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,6<----------------------0,3

             Fe + 2HCl --> FeCl₂ + H₂

            0,1<--0,2

=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)
PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

            \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)

=> \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄