\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\) \(\Rightarrow y=0,03\left(mol\right)\)

\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)

\(n_{H_2}=\dfrac{0,448}{22,4}=0,02mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,02                                  0,02   ( mol )

\(\Rightarrow x=0,02\left(mol\right)\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{0,02}{0,03}=\dfrac{2}{3}\)

\(\Rightarrow CTHH:Fe_2O_3\)

\(n_{H_2\left(thu\right)}=\dfrac{V}{22,4}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

1                                 :    1    (mol)

0,02                            :  0,02 (mol)

\(n_{H_2\left(dùng\right)}=\dfrac{V}{22,4}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

\(yH_2+Fe_xO_y\rightarrow^{t^0}xFe+yH_2O\)

y                      :       x  (mol)

0,03                 :      0,02 (mol)

\(\Rightarrow\dfrac{0,03}{y}=\dfrac{0,02}{x}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{0,02}{0,03}=\dfrac{2}{3}\Rightarrow x=2;y=3\)

-Vậy CTHH của oxit sắt là Fe₂O₃.

Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_xO_y}=a\left(mol\right)\end{matrix}\right.\)

=> 80a + 56ax + 16ay = 2,4 (1)

\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

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

                a---------------->ax

           Fe + 2HCl --> FeCl₂ + H₂

           ax--------------------->ax

=> \(ax=0,02\left(mol\right)\)

=> a = \(\dfrac{0,02}{x}\)

Thay vào (1)

\(80.\dfrac{0,02}{x}+56.0,02+\dfrac{16.0,02y}{x}=2,4\)

=> \(\dfrac{1,6}{x}+\dfrac{0,32y}{x}=1,28\)

=> 1,28x = 0,32y + 1,6

Chọn x = 2; y = 3 thỏa mãn

=> CTHH: Fe₂O₃

Fe₂O₃ + 3H₂ -> 2Fe + 3H₂O (1)

CuO + H₂ -> Cu + H₂O (2)

m_Fe2O3=72(g)\(\Leftrightarrow\)0,45(mol)

m_CuO=24(g)\(\Leftrightarrow\)0,3(mol)

Theo PTHH 1 ta có:

n_Fe=2n_Fe2O3=0,9(mol)

n_H2(1)=3n_Fe2O3=1,35(mol)

Theo PTHH 2 ta có:

n_Cu=n_CuO=n_H2(2)=0,3(mol)

m_Fe=56.1,35=75,6(g)

m_Cu=64.0,3=19,2(g)

V_H2=(1,35+0,3).22,4=36,96(lít)

Giả sử \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_M=1,5a\left(mol\right)\end{matrix}\right.\)

=> 27a + M_M.1,5a = 6,3 (g) (1)

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

- TH1: Nếu M không tác dụng với dd HCl

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,2<------------------0,3

=> a = 0,2 (mol)

(1) => M_M = 3 (L)

- TH2: Nếu M tác dụng với dd HCl

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             a----------------------->1,5a

            M + 2HCl --> MCl₂ + H₂

           1,5a---------------->1,5a

=> 1,5a + 1,5a = 0,3

=> a = 0,1

(1) => M_M = 24 (g/mol)

=> M là Mg

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{6,3}.100\%=42,857\%\\\%m_{Mg}=\dfrac{0,15.24}{6,3}.100\%=57,143\%\end{matrix}\right.\)

thank

\(n_{H_2}=0,65\left(mol\right)\)

\(n_{Cl_2}=0,75\left(mol\right)\)

\(Fe+2HCl-->FeCl_2+H_2\)

2x.................................................2x

\(2A+2nHCl-->2ACl_n+nH_2\)

3x....................................................3xn/2

\(2x+\dfrac{3xn}{2}=0,65\)

\(\Leftrightarrow4x+3xn=1,3\)

\(\Leftrightarrow x\left(4+3n\right)=1,3\)

\(\Leftrightarrow x=\dfrac{1,3}{4+3n}\left(1\right)\)

\(2Fe+3Cl_2-->2FeCl_3\)

2x........3x

\(2A+nCl_2-->2ACl_n\)

3x........3xn/2

\(3x+\dfrac{3xn}{2}=0,75\)

\(\Leftrightarrow6x+3xn=1,5\)

\(\Leftrightarrow3x\left(2+n\right)=1,5\)

\(\Leftrightarrow x\left(2+n\right)=0,5\left(2\right)\)

Thay (1) vào (2)

\(\Rightarrow n=3\)

vậy kim loại A có hóa tri là 3

X gồm Fe và Cu. Với HCl:
nFe = nH2 = 0,04
=>nCu = (mX – mFe)/64 = 0,02
=> nCuO = nFexOy = 0,02
-> x = nFe/nFexOy = 2
; Oxit là Fe2O3.

Bảo toàn O: \(m_{O\left(oxit\right)}=m_{giảm}=4,8-3,52=1,28\left(g\right)\)

\(n_{O\left(oxit\right)}=\dfrac{1,28}{16}=0,08\left(mol\right)\\ n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

           0,04 <------------------------ 0,02

\(m_{Cu}=3,52-0,04.56=1,28\left(g\right)\\ n_{O\left(CuO\right)}=n_{Cu}=\dfrac{1,28}{64}=0,02\left(mol\right)\\ n_{O\left(Fe_xO_y\right)}=0,08-0,02=0,06\left(mol\right)\)

CTHH: Fe_xO_y

=> x : y = 0,04 : 0,06 = 2 : 3

CTHH Fe₂O₃

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(n_{Fe}=n_{H_2}=0.2\left(mol\right)\)

\(m_{Cu}=m_{hh}-m_{Fe}=17.6-0.2\cdot56=6.4\left(g\right)\)

\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)

\(\Rightarrow m_{CuO}=0.1\cdot80=8\left(g\right)\)

\(m_{Fe_xO_y}=m_{hh}-m_{CuO}=24-8=16\left(g\right)\)

\(M_{Fe_xO_y}=\dfrac{16}{\dfrac{0.2}{x}}=80x\left(đvc\right)\)

\(\Leftrightarrow56x+16y=80x\)

\(\Leftrightarrow24x=16y\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{16}{24}=\dfrac{2}{3}\)

\(CT:Fe_2O_3\)

\(Fe + 2HCl \to FeCl_2 + H_2\\ n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2\ mol\\ \Rightarrow n_{Cu} = \dfrac{17,6-0,2.56}{64} = 0,1\ mol\)

BTNT với Fe,Cu

\(n_{CuO} = n_{Cu} = 0,1\ mol\\ n_{Fe_xO_y} = \dfrac{n_{Fe}}{x} = \dfrac{0,2}{x}mol\)

Suy ra ;

\(0,1.80 + \dfrac{0,2}{x}.(56x+16y) = 24\\ \Rightarrow \dfrac{x}{y} = \dfrac{2}{3}\)

Vậy oxit sắt cần tìm : Fe₂O₃

CTHH: Fe_xO_y

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,2<---------------------0,2

=> \(m_{Fe_xO_y}=12,8-0,2.56=1,6\left(g\right)\)

Trong 6,4g hỗn hợp rắn chứa 0,8g Fe_xO_y

\(n_{H_2O}=\dfrac{0,27}{18}=0,015\left(mol\right)\)

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

            \(\dfrac{0,015}{y}\)<---------------------0,015

=> \(M_{Fe_xO_y}=\dfrac{0,8}{\dfrac{0,015}{y}}=\dfrac{160}{3}y\)

=> \(56x+16y=\dfrac{160}{3}y\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow Fe_2O_3\)

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