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.b.

\(\left\{{}\begin{matrix}n_{Fe_2O_3}=40.80\%=32g\\m_{CuO}=40-32=8g\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{32}{160}=0,2mol\\n_{CuO}=\dfrac{8}{80}=0,1mol\end{matrix}\right.\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,1        0,1            0,1                 ( mol )

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

0,2            0,6               0,4                  ( mol )

\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68l\)

\(\left\{{}\begin{matrix}m_{Cu}=0,1.64=6,4g\\m_{Fe}=0,4.56=22,4g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{6,4}{6,4+22,4}.100=22,22\%\\\%m_{Fe}=100\%-22,22\%=77,78\%\end{matrix}\right.\)

c.
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) ( Cu không phản ứng với H₂SO₄ loãng )

0,4                                      0,4    ( mol )

\(V_{H_2}=0,4.22,4=8,96l\)

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

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

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

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

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

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

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

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

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

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

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

Đáp án:

 8,96 l

Giải thích các bước giải:

a)

Fe2O3+3H2->2Fe+3H2O

CuO+H2->Cu+H2O

gọi a là số mol Fe2O3 b là số mol CuO

Ta có

160a=2x80b=>a=b

ta có

112a+64b=17,6

a=b

=>a=0,1 b=0,1

nH2=0,1x3+0,1=0,4(mol)

VH2=0,4x22,4=8,96 l

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

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