Bài 7:
a)
mCu = 12,8 (g)
Gọi số mol Al, Zn là a, b (mol)
=> 27a + 65b = 58,8 - 12,8 = 46 (1)
\(n_{HCl}=0,625.4=2,5\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a---->3a-------->a------>1,5a
Zn + 2HCl --> ZnCl2 + H2
b--->2b------->b------->b
=> 3a + 2b = 2,5 (2)
(1)(2) => a = 0,5; b = 0,5
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{58,8}.100\%=22,96\%\\\%m_{Zn}=\dfrac{0,5.65}{58,8}.100\%=55,27\%\\\%m_{Cu}=\dfrac{12,8}{58,8}.100\%=21,77\%\end{matrix}\right.\)
b) \(n_{H_2}=1,5a+b=1,25\left(mol\right)\)
=> V = 1,25.22,4 = 28 (l)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,5.133,5=66,75\left(g\right)\\m_{ZnCl_2}=0,5.136=68\left(g\right)\end{matrix}\right.\)
c)
\(n_{AgCl}=n_{Cl}=2,5\left(mol\right)\)
=> mAgCl = 2,5.143,5 = 358,75 (g)
Bài 8:
a) mAg = 21,6 (g)
Gọi số mol Fe, Mg là a, b (mol)
\(n_{H_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
a--->2a-------->a----->a
Mg + 2HCl --> MgCl2 + H2
b---->2b--------->b---->b
=> a + b = 1,6
Và 127a + 95b = 177,6
=> a = 0,8; b = 0,8
\(\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{21,6}{21,6+0,8.56+0,8.24}.100\%=25,234\%\\\%m_{Fe}=\dfrac{0,8.56}{21,6+0,8.56+0,8.24}.100\%=52,336\%\\\%m_{Mg}=\dfrac{0,8.24}{21,6+0,8.56+0,8.24}.100\%=22,43\%\end{matrix}\right.\)
b) nHCl = 2a + 2b = 3,2 (mol)
=> \(aM=C_{M\left(ddHCl\right)}=\dfrac{3,2}{0,3}=10,67M\)
c) Kết tủa gồm \(\left\{{}\begin{matrix}Fe\left(OH\right)_2:0,8\left(mol\right)\\Mg\left(OH\right)_2:0,8\left(mol\right)\end{matrix}\right.\)
=> mkết tủa = 0,8.90 + 0,8.58 = 118,4 (g)
giúp mình với ạ ;-; . mình đang cần gấp lắm 
