PTHH:

(1) 2 Al + 6 HCl -> 2AlCl3 + 3 H2

x_____3x_________x_______1,5x (mol)

(2) Fe + 2 HCl -> FeCl2 + H2

y___________2y___y___y (mol)

Ta có: nH2= 13,44/22,4= 0,6(mol)

nH2(1) + nH2(2)= nH2(tổng)

<=> 1,5x+y=0,6 (a)

Ta có: mAl+mFe= 22,2

<=> 27x+56y=22,2 (b)

Từ (a), (b) ta có hpt:

\(\left\{{}\begin{matrix}1,5x+y=0,6\\27x+56y=22,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

Ta có: mAl= 0,2.27=5,4(g)

=> %mAl= \(\frac{5,4}{22,2}.100\approx24,324\%\)

=> \(\%mFe\approx100\%-24,324\%\approx75,676\%\)

* nAlCl3= x= 0,2(mol)

nFeCl2= y=0,3(mol)

=> mAlCl3= 133,5.0,2=26,7(g)

mFeCl2= 127.0,3= 38,1(g)

=> %mAlCl3= \(\frac{26,7}{26,7+38,1}.100\approx41,204\%\\ \Rightarrow\%mFeCl2\approx100\%-41,204\%\approx58,796\%\)

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

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

Ta có :

\(\left\{{}\begin{matrix}27x+56y=22,2\\1,5x+y=\frac{13,44}{22,4}\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

\(\rightarrow\%m_{Al}=\frac{0,2.27}{22,2}.100\%=24,32\%,\%m_{Fe}=100\%-24,32\%=75,68\%\)

\(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)

\(m_{FeCl_2}=0,3.127=38,1\left(g\right)\)

PTHH ( I ) : \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

......................x.........3x..............x............\(\frac{3}{2}\)x......

PTHH ( II ) : \(Fe+2HCl\rightarrow FeCl_2+H_2\)

.......................y.........2y..............y............y.......

- Gọi số mol của nhôm và sắt có trong hỗn hợp lần lượt là x, y (x, y> 0 )

a, - Ta có : \(m_{hh}=m_{Fe}+m_{Al}=n_{Fe}.M_{Fe}+n_{Al}.M_{Al}=22,2\)

=> \(56y+27x=22,2\) ( I )

\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{13,44}{22,4}=0,6\left(mol\right)\)

Mà \(n_{H_2}=n_{H_2\left(I\right)}+n_{H_2\left(II\right)}\)

=> \(1,5x+y=0,6\) ( II )

- Từ ( I ) và ( II ) ta có hệ phương trình : \(\left\{{}\begin{matrix}27x+56y=22,2\\1,5x+y=0,6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}27x+56\left(0,6-1,5x\right)=22,2\\y=0,6-1,5x\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}27x-84x=-11,4\\y=0,6-1,5x\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,2\\y=0,6-1,5.0,2=0,3\left(mol\right)\end{matrix}\right.\) ( TM )

=> \(\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

-> \(m_{Al}=n_{Al}.M_{Al}=0,2.27=5,4\left(g\right)\)

-> \(\%Al=\frac{m_{Al}}{m_{hh}}=\frac{5,4}{22,2}.100\%\approx24,3\%\)

Mà \(\%hh=\%Al+\%Fe=24,3\%+\%Fe=100\%\)

=> \(\%Fe=75,6\%\)

b, Ta có : \(\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{\left(Al\right)}=0,2\left(mol\right)\\n_{\left(Fe\right)}=0,3\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{AlCl_3}=0,2\left(mol\right)\\n_{FeCl_2}=0,3\left(mol\right)\end{matrix}\right.\)

=> m_{Muối khan} = \(m_{AlCl3}+m_{FeCl2}=n_{AlCl3}.M_{AlCl3}+n_{FeCl2}.M_{FeCl2}\)

=> m_{Muối khan} = \(0,2.\left(27+35,5.3\right)+0,3.\left(56+35,5.2\right)=64,8\left(g\right)\)

Chọn đáp án C

a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)

Chọn C

Đáp án C.

→ n C l - =   2 n H 2   =   0 , 16   → m m u o i   = 4 , 34   +   0 , 16 . 35 , 5   = 10 , 02