PTHH:

\(FeO+H_2\overset{t^o}{--->}Fe+H_2O\)

\(CuO+H_2\overset{t^o}{--->}Cu+H_2O\)

A: Fe, Cu

\(Fe+2HCl--->FeCl_2+H_2\)

\(Cu+HCl--\times-->\)

B: Cu

C: FeCl₂, HCl dư.

\(FeCl_2+2NaOH--->2NaCl+Fe\left(OH\right)_2\downarrow\)

\(HCl+NaOH--->NaCl+H_2O\)

D: Fe(OH)₂

a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{HCl}=0,5.0,4=0,2\left(mol\right)\)

PTHH: 2K + 2H₂O ---> 2KOH + H₂ (Fe và Cu ko tan trong nước)

              0,2                                0,1

Fe + 2HCl ---> FeCl₂ + H₂ (Cu ko phản ứng với HCl)

0,1     0,2

m_{Chất rắn còn lại} = m_Cu = 6,6 (g)

\(\rightarrow\left\{{}\begin{matrix}m_K=39.0,2=7,8\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\\m_{Cu}=6,6\left(g\right)\end{matrix}\right.\)

\(\rightarrow m_{\text{hhkimloại}}=7,8+5,6+6,6=20\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}\%m_K=\dfrac{7,8}{20}=39\%\\\%m_{Fe}=\dfrac{5,6}{20}=28\%\\\%m_{Cu}=100\%-39\%-28\%=33\%\end{matrix}\right.\)

b, PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

\(n_{O\left(\text{trong oxit}\right)}=n_{H_2O}=n_{H_2}=0,1\left(mol\right)\\ \rightarrow n_{Fe\left(\text{trong oxit}\right)}=\dfrac{5,8-0,1.16}{56}=0,075\left(mol\right)\)

\(\rightarrow x:y=n_{Fe}:n_O=0,075:0,1=3:4\)

CTHH của oxit sắt Fe₃O₄

Sửa đề thành 2,24 l khí C nhé :)

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

- Thấy Cu không phản ứng với HCl .

\(\Rightarrow m_{cr}=m_{Cu}=6,4\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

.x.......................................1,5x.........

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

.y....................................y.............

Theo bài ra ta có hệ : \(\left\{{}\begin{matrix}27x+56y+6,4=17,4\\1,5x+y=0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\) ( mol )

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=5,4\\m_{Fe}=5,6\end{matrix}\right.\) ( g )

b, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

.......0,1.........0,2...............................

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)

...0,2.......0,6..........................

\(\Rightarrow n_{NaOH}=0,2+0,6=0,8< 1\)

=> Trong B còn có HCl dư .

\(NaOH+HCl\rightarrow NaCl+H_2O\)

...0,2..........0,2....................

=> Dư 0,2 mol HCl .

\(\Rightarrow n_{HCl}=2n_{H_2}+0,2=1\left(mol\right)\)

\(\Rightarrow m_{ddB}=17,4+250-6,4-0,8=260,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{260,2}.100\%\approx2,8\%\\C\%_{FeCl_2}\approx4,88\%\\C\%_{AlCl_3}\approx10,26\%\end{matrix}\right.\)

Vậy ....