\(m_{CuO}=50.20\%=10\left(g\right)\)

\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)

\(m_{Fe_2O_3}=50-10=40\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\)

PTHH :

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

0,125   0,125   0,125 

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

0,25        0,75    0,5 

\(a,V_{H_2}=\left(0,75+0,125\right).22,4=19,6\left(l\right)\)

\(b,m_{Cu}=0,125.64=8\left(g\right)\)

\(m_{Fe}=0,5.56=28\left(g\right)\)

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 )

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

\(m_{CuO}=40.20\%=8\left(g\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{40-8}{160}=0,2\left(mol\right)\)

PTHH:

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

 0,1      0,1

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

0,2          0,6 

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

$m_{Fe_2O_3} = 60.80\% = 48(gam) \Rightarrow n_{Fe_2O_3} = \dfrac{48}{160} = 0,3(mol)$
$m_{CuO} = 60 - 48 = 12(gam) \Rightarrow n_{CuO} = \dfrac{12}{80} = 0,15(mol)$

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 1,05(mol)$
$V_{H_2} = 1,05.22,4 = 23,52(lít)$

VH2 = 23,52(lít) 

Ta có:

\(m_{Fe_2O_3}\) = m_hh. 80% = 25 . 80% = 20(g) => \(n_{Fe_2O_3}\) = \(\dfrac{m}{M}=\dfrac{20}{160}=0,125\left(mol\right)\)

=>\(m_{CuO}\) = m_hh - mFe₂O₃  = 25 - 20 = 5(g) => \(n_{CuO}\) = \(\dfrac{m}{M}=\dfrac{5}{80}=0,0625\left(mol\right)\)

Ta có: \(\Sigma n_O=3nFe_2O_3+nCuO=3.0,125+0,0625=0,4375\left(mol\right)\)

PT:

\(H_2+O\rightarrow H_2O\)

\(\Rightarrow n_O=n_{H_2}=0,4375\left(mol\right)\)

\(\Rightarrow V_{H_2}=n.22,4=0,4375.22,4=9,8\left(l\right)\) \(\Rightarrow B\)

Fe₂O₃+3H₂-to>2Fe+3H₂O

0,075----0,225

CuO+H₂-to>Cu+H₂O

0,1-----0,1

b)

m Fe2O3=20.\(\dfrac{60}{100}\)=12g

=>n Fe2O3=\(\dfrac{12}{160}\)=0,075 mol

m CuO=20-12=8g

=>n CuO=\(\dfrac{8}{80}=0,1mol\)

=>V_H2=(0,1+0,225).22,4=7,28l