a) \(2Al+6HCl-->2AlCl3+3H2\left(1\right)\)
x-----------------------------------------1,5x( mol)
\(Zn+2HCl-->ZnCl2+H2\)
y-------------------------------------y(mol)
\(n_{H2}=\frac{15,68}{22,4}=0,7\left(mol\right)\)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}27x+65y=31,4\\1,5x+y=0,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
\(\%m_{Al}=\frac{0,2.27}{31,4}.100\%=17,2\%\)
\(\%m_{Zn}=100-17,2=82,8\%\)
b) \(n_{HCl}=2n_{H2}=1,4\left(mol\right)\)
\(m_{HCl}=1,4.36,5=51,1\left(g\right)\)
\(m_{ddHCl}=\frac{51,1.100}{20}=255,5\left(g\right)\)
Do dùng dư 20%
--> \(m_{HCl}=255,5+255,5.20\%=306,6\left(g\right)\)
c) m dd sau pư = \(m_{KL}+m_{ddHCl}-m_{H2}=31,4+255,5-1,4\)
=\(288,3\left(g\right)\)
\(n_{Alcl3}=n_{Al}=0,2\left(mol\right)\)
\(m_{AlCl3}=0,2.122,5=24,5\left(g\right)\)
\(C\%_{Alcl3}=\frac{24,5}{288,3}.100\%=8,5\%\)
m\(_{ZnCl2}=0,4.136=54,4\left(g\right)\)
\(C\%_{ZnCl2}=\frac{54,4}{288,3}.100\%=18,87\%\)
\(31,4\left(g\right)\left\{{}\begin{matrix}Al\left(x\right)\\Z\left(y\right)\end{matrix}\right.+HCl\left(20\%\right)\rightarrow H_2\)
______________________________0,7
a, \(\left\{{}\begin{matrix}27x+65y=31,4\left(BTKL\right)\\3x+2y=0,7.2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right)\Rightarrow\%m_{Al}=17,2\%\)
\(m_{Zn}=0,4.65=26\left(g\right)\Rightarrow\%m_{Zn}=82,8\%\)
b,
\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)
0,2____0,6____0,2______0,3
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,4____0,8____0,4________0,4
\(n_{HCl}=0,14\Rightarrow m_{HCl}=5,11\left(g\right)\Rightarrow m_{HCl_{Can.dung}}=6,132\left(g\right)\)
c,
\(m_{dd}=31,4+6,132-0,7.2=36,132\left(g\right)\)
\(C\%_{AlCl3}=\frac{0,2.133,5}{36,132}=73,9\%\)
\(C\%_{ZnCl2}=\frac{0,4.\left(65+35,5.2\right)}{36,132}=15,1\%\)
