a) 

Gọi số mol Mg, Al là a, b (mol)

=> 24a + 27b = 26,25 (1)

\(n_{H_2}=\dfrac{30,8}{22,4}=1,375\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a-->2a--------->a------>a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

             b---->3b------->b------>1,5b

=> a + 1,5b = 1,375 (2)

(1)(2) => a = 0,25 (mol); b = 0,75 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{26,25}.100\%=22,857\%\\\%m_{Al}=\dfrac{0,75.27}{26,25}.100\%=77,143\%\end{matrix}\right.\)

b)

n_HCl = 2a + 3b = 2,75 (mol)

=> m_HCl = 2,75.36,5 = 100,375 (g)

=> \(m_{dd.HCl}=\dfrac{100,375.100}{10}=1003,75\left(g\right)\)

c) 

m_{dd sau pư} = 1003,75 + 26,25 - 1,375.2 = 1027,25 (g)

\(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,25.95}{1027,25}.100\%=2,312\%\\C\%_{AlCl_3}=\dfrac{0,75.133,5}{1027,25}.100\%=9,747\%\end{matrix}\right.\)

Gọi $n_{Fe} = a(mol) ; n_{Al} = b(mol)$

$\Rightarrow 56a + 27b = 11(1)$

$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = a + 1,5b = 8,96 : 22,4 = 0,4(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2

Vậy :

$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,91\%$
$\%m_{Al} = 100\% -50,91\% = 49,09\%$

Bài 1:

\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

PTHH: Fe + 2HCl → FeCl₂ + H₂

\(m_{H_2}=0,16.2=0,32\left(g\right)\)

\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)

Bài 12:

Theo ĐLBTKL, ta có:

\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

Chọn A

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

a, Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=1,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)

a) Bảo toàn khối lượng : $n_{O(oxit)} = \dfrac{23,88 - 15,72}{16} = 0,51(mol)$
$2H^+ + O^{2-} \to H_2O$
$n_{HCl} = 2n_O =  0,51.2 = 1,02(mol)$

b)

$n_{Cl^-} = n_{HCl} = 1,02(mol)$
Suy ra : 

$m_{muối} = m_{kim\ loại} + m_{Cl} = 15,72 + 1,02.35,5 =51,93(gam)$