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$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(m_{hh}=56a+24b=10.16\left(g\right)\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H_2}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.13,b=0.12\)
\(m_{Fe}=0.13\cdot56=7.28\left(g\right)\)
\(m_{Mg}=0.12\cdot24=2.88\left(g\right)\)
\(n_{HCl}=2\cdot n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,6}{2}>\dfrac{0,25}{1}\Rightarrow HCldư\\ Đặt:n_{Al}=t\left(mol\right);n_{Fe}=r\left(mol\right)\\ \left(t,r>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27t+56r=8,3\\1,5t+r=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=0,1\\r=0,1\end{matrix}\right.\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right);m_{Fe}=0,1.56=5,6\left(g\right)\\ b,n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ n_{Fe}=n_{FeCl_2}=0,1\left(mol\right)\Rightarrow m_{ddFeCl_2}=127.0,1=12,7\left(g\right)\\ m_{ddHCl}=300.1,15=345\left(g\right)\\ m_{ddsau}=8,3+345-0,25.2=352,8\left(g\right)\)
\(n_{HCl\left(dư\right)}=0,6-0,25.2=0,1\left(mol\right)\\ \Rightarrow m_{ddHCl}=0,1.36,5=3,65\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{352,8}.100\approx1,035\%\\ C\%_{ddAlCl_3}=\dfrac{13,35}{352,8}.100\approx3,784\%\\ C\%_{ddFeCl_2}=\dfrac{12,7}{352,8}.100\approx3,6\%\)
a) Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 8,56 (1)
\(n_{H_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a--->2a-------->a----->a
Fe + 2HCl --> FeCl2 + H2
b----->2b------->b------>b
=> a + b = 0,14 (2)
(1)(2) => a = 0,08; b = 0,06
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,08.65}{8,56}.100\%=60,748\%\\\%m_{Fe}=\dfrac{0,06.56}{8,56}.100\%=39,252\%\end{matrix}\right.\)
b)
nKOH = 0,2.0,1 = 0,02 (mol)
PTHH: KOH + HCl --> KCl + H2O
0,02-->0,02
=> nHCl = 0,02 + 2a + 2b = 0,3 (mol)
=> \(C_{M\left(HCl\right)}=xM=\dfrac{0,3}{0,15}=2M\)
c) m = 0,08.136 + 0,06.127 = 18,5(g)
a) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
mCu = mY = 9,6 (g)
Gọi số mol Al, Mg là a, b
=> 27a + 24b = 14,7 - 9,6 = 5,1 (g)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a-->3a-------->a------>1,5a
Mg + 2HCl --> MgCl2 + H2
b--->2b------->b----->b
=> 1,5a + b = 0,25
=> a = 0,1; b = 0,1
=> \(\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Cu}=9,6\left(g\right)\end{matrix}\right.\)
b) nHCl(PTHH) = 3a + 2b = 0,5 (mol)
=> nHCl(thực tế) = \(\dfrac{0,5.120}{100}=0,6\left(mol\right)\)
=> \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3M\)
c) \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(HCldư\right)}=\dfrac{0,6-0,5}{0,2}=0,5M\end{matrix}\right.\)
d) \(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)
PTHH: \(Cu+Cl_2\underrightarrow{t^o}CuCl_2\)
0,15-->0,15
=> \(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)
\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)