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Gọi nFe = a (mol); nMg = b (mol)
56a + 24b = 8 (1)
nH2 = 4,48/22,4 = 0,2 (mol)
PTHH:
Fe + 2HCl -> FeCl2 + H2
a ---> a ---> a ---> a
Mg + 2HCl -> MgCl2 + H2
b ---> b ---> b ---> b
a + b = 0,2 (2)
(1)(2) => a = b = 0,1 (mol)
mFe = 0,1 . 56 = 5,6 (g)
%mFe = 5,6/8 = 70%
%mMg = 100% - 70% = 30%
nHCl = 0,1 . 2 + 0,1 . 2 = 0,4 (mol)
CMddHCl = 0,4/0,1 = 4M
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
Mg+2HCl->MgCl2+H2
x------2x--------x---------x
2Al+6HCl->2AlCl3+3H2
y---------3y-----y--------3\2y
ta có :
\(\left\{{}\begin{matrix}24x+27y=11,7\\x+\dfrac{3}{2}y=0,6\end{matrix}\right.\)
=>x=0,15 mol, y=0,3 mol
=>%mMg=\(\dfrac{0,15.24}{11,7}.100=30,77\%\)
=>%mAl=100-30,77=69,23%
b)
m HCl=1,2.36,5=43,8g
=>C%=\(\dfrac{43,8}{200}.100\)=21,9%
a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)
