a, Ta có: 27n_Al + 56n_Fe = 22 (1)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Fe + 2HCl ---> FeCl₂ + H₂

0,15    0,3           0,15     0,15

m_Fe = 0,15.56 = 8,4 (g)

m_Fe2O3 = 24,4 - 8,4 = 16 (g)

n_Fe2O3 = \(\dfrac{16}{160}=0,1\left(mol\right)\)

%m_Fe = \(\dfrac{8,4}{24,4}=34,42\%\)

%m_Fe2O3 = \(100\%-34,42\%=65,58\%\)

Fe₂O₃ + 6HCl ---> 2FeCl₃ + 3H₂O

0,1            0,6          0,2           0,3

n_{HCl (ban đầu)} = 0,8.1,5 = 1,2 (mol)

n_{HCl (dư)} = 1,2 - 0,3 - 0,6 = 0,3 (mol)

=> \(\left\{{}\begin{matrix}C_{MFeCl_3}=\dfrac{0,2+0,15}{0,8}=0,4375M\\C_{MHCl\left(dư\right)}=\dfrac{0,3}{0,8}=0,375M\end{matrix}\right.\)

PTHH: 

FeCl₃ + 3NaOH ---> Fe(OH)₃ + 3NaCl

0,35        1,05

HCl + NaOH ---> NaCl + H₂O

0,3       0,3

=> \(V_{ddNaOH}=\dfrac{1,05+0,3}{1}=1,35\left(l\right)=1350\left(ml\right)\)

a) PTHH: Fe + 2HCl ===> FeCl₂ + H₂

MgO + 2HCl ===> MgCl₂ + H₂O

b) Ta có: n_H2 =\(\frac{2,24}{22,4}=0,1\left(mol\right)\)

=> n_Fe = n_H2 = 0,1 (mol)

=> m_Fe = 0,1 x 56 = 5,6 gam

=> m_MgO = 22,4 - 5,6 = 156,8 gam

c. %_MgO = \(\frac{16,8}{22,4}.100\%=75\%\)

=> %_Fe = 100% - 75% = 25%

a)

Gọi : \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)⇒ 27a + 56b = 1,66(1)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe +2 HCl \to FeCl_2 + H_2\)

Theo PTHH :

\(n_{H_2} = 1,5a + b = \dfrac{1,12}{22,4} = 0,05(2)\)

Từ (1)(2) suy ra a = 0,02 ; b = 0,02

Vậy :

\(\%m_{Al} = \dfrac{0,02.27}{1,66}.100\% = 32,53\%\\ \%m_{Fe} = 100\% - 32,53\% = 67,47\%\)

a)

\(n_{HCl} = 2n_{H_2} = 0,05.2 = 0,1(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,1.36,5}{100}.100\% = 3,65\%\)

nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)

Chọn đáp án C

Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 5,5 (1)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,5}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)