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\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,6}{2}>\dfrac{0,25}{1}\Rightarrow HCldư\\ Đặt:n_{Al}=t\left(mol\right);n_{Fe}=r\left(mol\right)\\ \left(t,r>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27t+56r=8,3\\1,5t+r=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=0,1\\r=0,1\end{matrix}\right.\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right);m_{Fe}=0,1.56=5,6\left(g\right)\\ b,n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ n_{Fe}=n_{FeCl_2}=0,1\left(mol\right)\Rightarrow m_{ddFeCl_2}=127.0,1=12,7\left(g\right)\\ m_{ddHCl}=300.1,15=345\left(g\right)\\ m_{ddsau}=8,3+345-0,25.2=352,8\left(g\right)\)
\(n_{HCl\left(dư\right)}=0,6-0,25.2=0,1\left(mol\right)\\ \Rightarrow m_{ddHCl}=0,1.36,5=3,65\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{352,8}.100\approx1,035\%\\ C\%_{ddAlCl_3}=\dfrac{13,35}{352,8}.100\approx3,784\%\\ C\%_{ddFeCl_2}=\dfrac{12,7}{352,8}.100\approx3,6\%\)
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: x 1,5x
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)
b)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,1 0,1 0,1
\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)
mdd sau pứ = 5,1+250-0,15.2 = 254,8(g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)
\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)
Ta có: \(m_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.pư\right)}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=309,6\left(g\right)\)
\(\Rightarrow a=309,6-300=9,6\left(g\right)\)
\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)
a)
Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$
$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$
Từ (1)(2) suy ra : a = 0,1; b = 0,2
$m_{Zn} = 0,1.65 = 6,5(gam)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$
a.\(n_{H_2}=\dfrac{7,28}{22,4}=0,325mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Zn}=y\end{matrix}\right.\) \(\left(mol\right)\) \(\rightarrow27x+65y=10,55\left(g\right)\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
x 1/2 x 3/2 x ( mol )
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
y y y ( mol )
\(\rightarrow\dfrac{3}{2}x+y=0,325\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,15.27}{10,55}.100\%=38,38\%\\\%m_{Zn}=100\%-38,38\%=61,62\%\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,15=0,075\\n_{ZnSO_4}=0,1\end{matrix}\right.\) ( mol )
\(\left\{{}\begin{matrix}C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,075}{0,8}=0,09M\\C_{M_{ZnSO_4}}=\dfrac{0,1}{0,8}=0,125M\end{matrix}\right.\)