a)

CuO + H₂ --t^o--> Cu + H₂O

Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

b) \(n_{Fe_2O_3}=\dfrac{32.20\%}{160}=0,04\left(mol\right)\)

\(n_{CuO}=\dfrac{32-0,04.160}{80}=0,32\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

            0,32-->0,32---->0,32

            Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

             0,04-->0,12-------->0,08

=> V_H2 = (0,32 + 0,12).22,4 = 9,856 (l)

c)

m_Cu = 0,32.64 = 20,48 (g)

m_Fe = 0,08.56 = 4,48 (g)

\(a,\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)

Loại phản ứng: Phản ứng thế

\(b,n_{Fe}=2.n_{Fe_2O_3}+3.n_{Fe_3O_4}=2.\dfrac{32}{160}+3.0,15=0,85\left(mol\right)\\ m_{Fe}=0,85.56=47,6\left(g\right)\\ c,n_{H_2}=\dfrac{32}{160}.3+4.0,15=1,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,2.22,4=28\left(l\right)\)

Em xem sao oxit sắt lại hỏi KL nhôm nha! Vô lí!!!

a, -Gọi số mol của CuO và Fe2O3 lần lượt là x, y ( mol )

PTKL : \(80x+160y=40\left(I\right)\)

\(CuO+H_2\rightarrow Cu+H_2O\)

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

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

...y............3y......

=> \(n_{H_2}=x+3y=\dfrac{V}{22,4}=0,6\left(mol\right)\left(II\right)\)

- Giair I và II ta được : x = 0,3 , y = 0,1 ( mol )

=> \(\left\{{}\begin{matrix}mCuO=n.M=24\left(g\right)\\mFe2O3=mhh-mCuO=16\left(g\right)\end{matrix}\right.\)

b, \(\%CuO=\dfrac{m}{mhh}.100\%=60\%\)

=> %Fe2O3 =100% - %CuO = 40% .

Vậy ...

Bài 1:

a) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+2H_2O\)

                \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20\cdot80\%=16\left(g\right)\\m_{CuO}=20-16=4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,35\left(mol\right)\) \(\Rightarrow V_{H_2}=0,35\cdot22,4=7,84\left(l\right)\)

c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu}=n_{CuO}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{hhB}=m_{Fe}+m_{Cu}=0,2\cdot56+0,05\cdot64=14,4\left(g\right)\)

Bài 2:

PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

            \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

a) Vì khối lượng Cu bằng \(\dfrac{6}{5}\) khối lượng Fe 

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=\dfrac{26,4}{6+5}\cdot6=14,4\left(g\right)\\m_{Fe}=26,4-14,4=12\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\\n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\) \(\Rightarrow V_{H_2}=\dfrac{153}{280}\cdot22,4=12,24\left(l\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{28}\left(mol\right)\\n_{CuO}=n_{Cu}=0,225\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=\dfrac{3}{28}\cdot160\approx17,14\left(g\right)\\m_{CuO}=0,225\cdot80=18\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=35,14\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{17,14}{35,14}\cdot100\%\approx48,78\%\\\%m_{CuO}=51,22\%\end{matrix}\right.\)

Bài 3:

PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\uparrow\)

            \(Fe_3O_4+4CO\xrightarrow[]{t^o}3Fe+4CO_2\uparrow\)

a) Ta có: \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{31,2+15,2}{2}=23,2\left(g\right)\\m_{CuO}=31,2-23,2=8\left(g\right)\end{matrix}\right.\)

\(\Rightarrow n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)=\dfrac{8}{80}=n_{CuO}\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,3\left(mol\right)\\n_{Cu}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\m_{Cu}=0,1\cdot64=6,4\left(g\right)\end{matrix}\right.\)

b) Theo các PTHH: \(n_{CO}=n_{CO_2}=n_{Cu}+\dfrac{4}{3}n_{Fe}=\dfrac{7}{30}\left(mol\right)\)

\(\Rightarrow V_{CO}=V_{CO_2}=\dfrac{7}{30}\cdot22,4\approx5,23\left(l\right)\)  

\(\left\{{}\begin{matrix}CuO:a\\Fe2O3:2a\end{matrix}\right.\)

a.\(80a+320a=24\Leftrightarrow a=0.06\)

\(\Rightarrow\left\{{}\begin{matrix}CuO=0.06\\Fe2O3=0.12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}CuO=4.8g\\Fe2O3=19.2g\end{matrix}\right.\)

b.\(CuO+H2\rightarrow Cu+H2O\)

    a           a         a

\(Fe2O3+3H2\rightarrow2Fe+3H2O\)

    2a          6a       4a

\(\Rightarrow V_{H2}=\left(a+6a\right)\times22.4=9.408l\)

c.nHCl = 0.2 mol

\(Fe+2HCl\rightarrow FeCl2+H2\)

 0.1     0.2

m chất rắn còn lại = mCu + m Fe ban đầu - m Fe bị hòa tan

                             = \(a\times64+4a\times56-0.1\times56=11.68g\)

2 kim loại là Fe và Cu; Cu không phản ứng với H2SO4 loãng.

a) Số mol Fe = số mol H2 = 5,04/22,4 = 0,225 mol; số mol Fe2O3 = 1/2 số mol Fe = 0,1125 mol.

mFe2O3 = 160.0,1125 = 18 g; mCuO = 24-18 = 6 g; %Fe2O3 = 18/24 = 75%; %CuO = 25%.

b) Số mol H2 = số mol H2O = số mol O = 3nFe2O3 + nCuO = 3.0,1125 + 6/80 = 0.4125 mol. V = 9,24 lít.

a)

\(m_{CuO}=\dfrac{32.40}{100}=12,8\left(g\right)\) => \(n_{CuO}=\dfrac{12,8}{80}=0,16\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{32-12,8}{160}=0,12\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

           0,16->0,16---->0,16

             Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

              0,12-->0,36----->0,24

=> \(V_{H_2}=\left(0,16+0,36\right).22,4=11,648\left(l\right)\)

b)

m_Cu = 0,16.64 =10,24 (g)

m_Fe = 0,24.56 = 13,44 (g)

c) 

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

 Xét tỉ lệ: \(\dfrac{0,24}{1}< \dfrac{0,5}{2}\) => HCl dư, Fe hết

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,24------------------->0,24

=> \(V_{H_2}=0,24.22,4=5,376\left(l\right)\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=0,5a\left(mol\right)\\ m_{hhB}=17,6\\ \Leftrightarrow56a+64.0,5a=17,6\\ \Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow n_{Fe}=0,2\left(mol\right);n_{Cu}=0,1\left(mol\right)\\ a,n_{H_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,2+0,1=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ b,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ \Rightarrow ddC:FeCl_2,HCldư\\ n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

PTHH $F{e}_2{O}_3+3H_2--->2Fe+3H_{2}O$
_________x____________________3x

$FeO+H_2--->Fe+H_{2}O$
___y_____________________y
Gọi x, y lần lượt là số mol $F{e}_2{O}_3$ và $FeO$
$n_{H_{2}O}= \backslash (\backslash frac\{2,88\}\{18\}\backslash )=0,16mol$
Ta có $160x+72y=9,6$
$3x+y=0,16$
$\mathrm{<=>x}=0,03mol;y=0,06mol$

a) Gọi số mol H₂ là x

=> nH2O=x(mol)

Theo ĐLBTKL: mA+mH2=mB+mH2O

=> 200 + 2x = 156 + 18x

=> x = 2,75 (mol)

=> VH2=2,75.22,4=61,6(l)

b) Gọi nCuO=a(mol)

nFe2O3=1,5a(mol)

=> 80a + 240a + 102b = 200

=> 320a + 102b = 200

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              a---------------->a

            Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

             1,5a------------------>3a

=> 64a + 168a + 102b = 156

=> 232a + 102b = 156

=> a = 0,5; b = \(\dfrac{20}{15}\)

%mCuO=\(\dfrac{0,5.80}{200}\).100%=20%

%mFe2O3=\(\dfrac{0,75.160}{200}\).100%=60%

%mAl2O3=\(\dfrac{\dfrac{20}{15}102}{200}\).100%=20%

c) nH2=\(\dfrac{2,75}{5}\)=0,55(mol)

nFeO(tt)=\(\dfrac{36}{72}\)=0,5(mol)

Gọi số mol FeO phản ứng là t (mol)

PTHH: FeO + H₂ --t^o--> Fe + H₂O

              t--------------->t

=> 56t + (0,5-t).72 = 29,6

=> t = 0,4 (mol)

=> H%=\(\dfrac{0,4}{0,5}\).100%=80%

tỉ leejlaf 1:1,5 chứ