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 )

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

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

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

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

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

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

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

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

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

\(m_{Fe_2O_3}=80\%.50=40\left(g\right)\Rightarrow n_{Fe_2O_3}=0,25\left(mol\right)\\ m_{CuO}=50-40=10\left(g\right)\Rightarrow n_{CuO}=0,125\left(mol\right)\\Fe_2O_3+3H_2-^{t^o}\rightarrow 2Fe+3H_2O\\ CuO+H_2-^{t^o}\rightarrow Cu+H_2O\\ \Sigma n_{H_2}=0,25.3+0,125=0,875\left(mol\right)\\ \Rightarrow V_{H_2}=0,875.22,4=19,6\left(l\right)\)

\(m_{Fe_2O_3}=\dfrac{80\cdot50}{100}=40\left(g\right)\)

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

\(n_{Fe_2O_3}=\dfrac{40}{160}=0.25\left(mol\right)\)

\(n_{CuO}=\dfrac{10}{80}=0.125\left(mol\right)\)

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

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

\(n_{H_2}=3\cdot0.25+0.125=0.875\left(mol\right)\)

\(V_{H_2}=0.875\cdot22.4=19.6\left(l\right)\)

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Đáp án đúng : A