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 : \(\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{\left(Al\right)}=0,2\left(mol\right)\\n_{\left(Fe\right)}=0,3\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{AlCl_3}=0,2\left(mol\right)\\n_{FeCl_2}=0,3\left(mol\right)\end{matrix}\right.\)
=> mMuối khan = \(m_{AlCl3}+m_{FeCl2}=n_{AlCl3}.M_{AlCl3}+n_{FeCl2}.M_{FeCl2}\)
=> mMuối khan = \(0,2.\left(27+35,5.3\right)+0,3.\left(56+35,5.2\right)=64,8\left(g\right)\)
