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

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

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

MgO + 2HCl -> MgCl2 + H2O

x            2x

Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

 y             6y

CuO + 2HCl -> CuCl2 + H2O

z           2z

A--H2 dư----> MgO          + H2O

                         Fe

                         Cu

Fe2O3 + 3H2 -> 2Fe + 3H2O

ky

CuO + H2 -> Cu + H2O

kz

nH2O=1,62/18=0,09mol

Gọi nMgO=x, nFe2O3=y, nCuO=z (trong 4,8g hh A)

      nMgO=Kx, nFe2O3=Ky, nCuO=Kz (trong 0,09mol A)

=> kx+ky+kz =4,8

      3ky + kz= 0,09

=> 2ky = kx => 2y=x

                          40x + 160y +80z=4,8

                           nHCl=2x+6y+2z=5,84/36,5=0,16

=> x=0,02            => mMgO= 0,8g

      y=0,01                  mFe2O3=1,6g

      z=0,03                  mCuO=2,4g

*tk

Đặt \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\\n_{Fe_2O_3}=c\left(mol\right)\end{matrix}\right.\) \(\Rightarrow102a+40b+160c=2,22\)

\(Al_2O_3,MgO\) không bị khử bởi \(CO\)

\(PTHH:Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

\(\left(mol\right)\)          \(c\)                     \(2c\)

\(\Rightarrow102a+40b+56.2c=1,98\)

\(PTHH:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(\left(mol\right)\)          \(a\)          \(6a\)

\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(\left(mol\right)\)        \(b\)            \(2b\)

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

\(\left(mol\right)\)        \(2c\)        \(4c\)

\(\Rightarrow6a+2b+4c=0,1\)

Từ đó: \(\left\{{}\begin{matrix}a=0,01\left(mol\right)\\b=0,01\left(mol\right)\\c=0,005\left(mol\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%m_{Al_2O_3}=45,95\left(\%\right)\\\%m_{MgO}=18,02\left(\%\right)\\\%m_{Fe_2O_3}=36,03\left(\%\right)\end{matrix}\right.\)

3. Đặt n_Al2O3= x, n_MgO=y, n_Fe2O3 = z
PTHH:
Fe2O3 + 3CO-----> 2Fe + 3CO2
Al2O3 + 6HCl ------> 2AlCl3 + 3H2O
MgO + 2HCl --------> MgCl2 + H2O
Fe + 2HCl -------> FeCl2 + H2

Ta có khối lượng của hỗn hợp X : \(102x+40y+160z=2,22\) (1)

Chất rắn Y gồm Fe, MgO và Al2O3

=> \(56.2z+102x+40y=1,98\) (2)

Theo PT ta có : \(n_{HCl}=6x+2y+2z.2=0.1.1\) (3)

Từ (1), (2), (3) => x=0,01 ; y=0,01, z= 0,005

=> \(\%m_{Al_2O_3}=\dfrac{0,01.102}{2,22}.100=45,95\%\)

\(\%m_{MgO}=\dfrac{0,01.102}{2,22}.100=18,01\%\)

=>\(\%m_{Fe_2O_3}=\dfrac{0,005.160}{2,22}.100=36,04\%\)

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

Gọi \(\left\{{}\begin{matrix}n_M=a\left(mol\right)\\n_{Al}=2a\left(mol\right)\end{matrix}\right.\)

=> a.M_M + 54a = 15,6 (1)

\(n_{Cl_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

M⁰ - 2e --> M⁺²

a--->2a

Al⁰ - 3e --> Al⁺³

2a-->6a

Cl₂⁰ + 2e --> 2Cl^-1

0,6-->1,2

2H⁺¹ + 2e --> H₂⁰

            0,4<--0,2

Bảo toàn e: 2a + 6a = 1,6 

=> a = 0,2 

Thay vào (1) => M_M = 24 (g/mol)

=> M là Mg

b) Xét \(m_{Mg}+m_{Al}=0,1.24+0,2.27=7,8\left(g\right)\)

=> Không có khí thoát ra

=> pư tạo ra sản phẩm khử là NH₄NO₃

PTHH: 4Mg + 10HNO₃ --> 4Mg(NO₃)₂ + NH₄NO₃ + 3H₂O

             0,1---->0,25

            8Al + 30HNO₃ --> 8Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O

            0,2--->0,75 

=> n_HNO3 = 0,25 + 0,75 = 1 (mol)

a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

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

Theo PT: \(n_{Fe}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Cu}=24-m_{Fe}=12,8\left(g\right)\) \(\Rightarrow n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m = m_CuO + m_Fe2O3 = 0,2.80 + 0,1.160 = 32 (g)

b, \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,2.80}{32}.100\%=50\%\\\%m_{Fe_2O_3}=50\%\end{matrix}\right.\)

c, Theo PT: \(n_{H_2}=n_{CuO}+3n_{Fe_2O_3}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)

reo?

\(n_{CuO}=2a\left(mol\right)\Rightarrow n_{Fe_2O_3}=a\left(mol\right)\)

\(m_X=80\cdot2a+160a=80\left(g\right)\)

\(\Rightarrow a=0.25\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(n_{H_2}=0.5+0.25\cdot3=1.25\left(mol\right)\)

\(V_{H_2}=1.25\cdot22.4=28\left(l\right)\)

\(m_{cr}=0.5\cdot64+0.5\cdot56=60\left(g\right)\)