Ta có:

\(\frac{m_{Fe2O3}}{m_{CuO}}=\frac{2}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}3m_{Fe2O3}-2m_{CuO}=0\\480m_{Fe2O3}-160n_{CuO}=0\\160n_{Fe2O3}+80n_{CuO}=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Fe2O3}=0,08\left(mol\right)\\n_{CuO}=0,24\left(mol\right)\end{matrix}\right.\)

PTHH:

\(Fe_2O_3+3H_2\rightarrow3Fe+3H_2O\)

0,08____0,24______0,16__________

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

0,24___0,24____0,24_____

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,16.56}{0,16.56+0,24.64}.100\%=36,84\%\\\%m_{Cu}=100\%-36,84\%=63,16\%\end{matrix}\right.\)

\(\Rightarrow V_{H2}=0,48.22,4=10,752\left(l\right)\)

Đặt \(n_{Fe_2O_3}=x\left(mol\right);n_{CuO}=y\left(mol\right)\)

Theo đề bài ta có:

\(\left\{{}\begin{matrix}160x+80y=32\\\frac{x}{y}=\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{35}\\y=\frac{6}{35}\end{matrix}\right.\)

\(\%m_{Fe_2O_3}=\frac{\frac{160.4}{35}}{32}.100\%=57,1\left(\%\right)\Rightarrow\%m_{CuO}=100-57,1=42,9\left(\%\right)\)

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

(mol)_____\(\frac{4}{35}\)______\(\frac{12}{35}\)____________

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

(mol)____\(\frac{6}{35}\)_____\(\frac{6}{35}\)__________

\(V_{H_2}=22,4.\left(\frac{6}{35}+\frac{12}{35}\right)=11,52\left(l\right)\)

Tính lại cái %m nhé, tại chưa đọc kĩ đề :)

\(\%m_{Fe}=\frac{\frac{56.8}{35}}{\frac{56.8}{35}+\frac{80.6}{35}}.100\%=48,3\left(\%\right)\Rightarrow\%m_{Cu}=100-48,3=51,7\left(\%\right)\)

đồng 3 oxit á ý c ik 

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

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

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

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

PTHH: CuO + H₂ → Cu + H₂O

Mol:      x         x         x

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:       y            3y        2y

Ta có hpt:\(\left\{{}\begin{matrix}80x+160y=14\\x+3y=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(m_{hh.kim.loại}=m_{Cu}+m_{Fe}=0,075.64+2.0,05.56=10,4\left(g\right)\)

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

PTHH:

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

Theo 2 pthh trên: \(n_{H_2O}=n_{H_2}=0,225\left(mol\right)\)

\(\rightarrow m_{H_2O}=0,225.18=4,05\left(g\right)\\ \rightarrow m_{H_2}=0,225.2=0,45\left(g\right)\)

Áp dụng ĐLBTKL, ta có:

\(m_{oxit\left(CuO,Fe_2O_3\right)}+m_{H_2}=m_{\text{kim loại}\left(Cu,Fe\right)}+m_{H_2O}\\ \rightarrow m_{\text{kim loại}\left(Cu,Fe\right)}=14+0,45-4,05=10,4\left(g\right)\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01mol\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)

\(PbO+H_2\rightarrow\left(t^o\right)Pb+H_2O\)

0,01    0,01            0,01            ( mol )

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

0,025  0,025        0,025            ( mol )

\(m_{hh}=m_{PbO}+m_{CuO}=\left(0,01.223\right)+\left(0,025.80\right)=4,23g\)

\(V_{H_2}=\left(0,01+0,025\right).22,4=0,784l\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01\left(mol\right)\\ n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\\ PTHH:PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ Mol:0,01\leftarrow0,01\leftarrow0,01\\ CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,025\leftarrow0,025\leftarrow0,025\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,025.80=2\left(g\right)\\m_{PbO}=0,01.223=2,23\left(g\right)\end{matrix}\right.\Rightarrow m_{oxit}=2+2,23=4,23\left(g\right)\\ V_{H_2}=\left(0,01+0,025\right).22,4=0,784\left(l\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)

Theo đề gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=3x\left(mol\right)\\n_{CuO}=2x\left(mol\right)\end{matrix}\right.\)

Có: \(m_{hh}=m_{Fe_2O_3}+m_{CuO}=160.3x+80.2x=32\)

\(\Rightarrow x=0,05\\ \Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,05.3=0,15\left(mol\right)\\n_{CuO}=0,05.2=0,1\left(mol\right)\end{matrix}\right.\)

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

0,15 ---->0,45-->0,3

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

0,1 --->0,1-->0,1

a. \(m_{kim.loại}=m_{Fe}+m_{Cu}=0,3.56+0,1.64=23,2\left(g\right)\)

b. \(V_{H_2}=\left(0,45+0,1\right).22,4=12,32\left(l\right)\)

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

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