Gọi x,y lần lượt là số mol của Al, Fe

n_H2 = \(\dfrac{8,96}{22,4}\)=0,4 mol

Pt: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

......x.................................0,5x...........1,5x

.....Fe + H₂SO₄ --> FeSO₄ + H₂

.......y..........................y............y

Ta có hệ pt:

 {27x+56y=11

   1,5x+y=0,4

⇔x=0,2, y=0,1

% m_Al = \(\dfrac{0,2.27}{11}\).100%=49,1%

% m_Fe = \(\dfrac{0,1.56}{11}\).100%=50,9%

m_Al2(SO4)3 = 0,5x . 342 = 0,5 . 0,2 . 342 = 34,2 (g)

m_FeSO4 = 152y = 152 . 0,1 = 15,2 (g)

Gọi CTTQ: M_xO_y

Pt: M_xO_y + yH₂ --to--> xM + yH₂O

\(\dfrac{0,4}{y}\)<-------0,4

Ta có: 232,2=\(\dfrac{0,4}{y}\)(56x+16y)

⇔23,2=\(\dfrac{22,4x}{y}\)+6,4

⇔\(\dfrac{22,4x}{y}\)=16,8

⇔22,4x=16,8y

⇔x:y=3:4

Vậy CTHH của oxit: Fe₃O₄

\(m_{cr}=m_{Fe}=12.8\left(g\right)\)

\(NaOH+Al+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)

\(m_{Al}=m_{hh}-m_{Fe}=18.2-12.8=5.4\left(g\right)\)

\(\%m_{Al}=\dfrac{5.4}{18.2}\cdot100\%=29.67\%\)

Al pư NaOH, Fe không pư NaOH nhé, nên chất rắn sau pư là Fe

\(2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2\)

\(m_{Al}= m_{hh} - m_{Fe}= 18,2 - 12,8 = 5,4 g\)

%m_Al=\(\dfrac{5,4}{18,2} . 100\)%  = 29,67%

PTHH: \(MgO+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O\)  (1)

       \(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\)  (2)

Ta có: \(n_{NO}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\) \(\Rightarrow n_{Mg}=0,0225\left(mol\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{0,0225\cdot24}{9,4}\cdot100\%\approx5,74\%\) \(\Rightarrow\%m_{MgO}=94,26\%\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{Mg\left(NO_3\right)_2\left(2\right)}=0,0225mol\\n_{Mg\left(NO_3\right)_2\left(1\right)}=n_{MgO}=\dfrac{9,4-m_{Mg}}{40}=0,2215\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{Mg\left(NO_3\right)_2}=0,244\left(mol\right)\) \(\Rightarrow m_{Mg\left(NO_3\right)_2}=0,244\cdot148=36,112\left(g\right)\)

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

0,6--------------------------0,9

Al2O3+6HCl-> 2AlCl3+3H2O

n H2=0,9 mol

=>m Al=0,6.27=16,2g

=>%mAl=\(\dfrac{16,2}{36,6}100\)=44,26%

=>%m Al2O3=55,74%

\(n_{H_2}=\dfrac{20,16}{22,4}=0,9mol\)

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

0,6                                     0,9

\(m_{Al}=0,6\cdot27=16,2g\)

\(\%m_{Al}=\dfrac{16,2}{36,6}\cdot100\%=44,26\%\)

\(\%m_{Al_2O_3}=100\%-44,26\%=55,73\%\)

\(n_{H_2}=\dfrac{20,16}{22,4}=0,9mol\)

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

 0,6                                       0,9    ( mol )

( \(Al_2O_3+HCl\) không giải phóng \(H_2\) )

\(\rightarrow m_{Al}=0,6.27=16,2g\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{16,2}{36,6}.100=44,26\%\\\%m_{Al_2O_3}=100\%-44,26\%=55,74\%\end{matrix}\right.\)

Chọn A

Đáp án A

n H 2   =   0 , 2   ( m o l )

=> m_hh= m_Fe + m_Al

Bảo toàn electron:

Mg + 2HCl → MgCl₂  + H₂ (1)

Al₂O₃ + 6HCl  →  2AlCl₃  + 3H₂O (2)

nH₂ = 2,8/22,4 = 0,125 mol 

Theo tỉ lệ phản ứng (1) => nMg = nH₂ = 0,125 mol

<=> mMg = 0,125 .24 = 3 gam và mAl₂O₃ = 8,1 - 3 =5,1 gam

%mMg = \(\dfrac{3}{8,1}\).100% = 37,03% => %mAl₂O₃ = 100 - 37,03 = 62,97%

b) nAl₂O₃ = \(\dfrac{5,1}{102}\)= 0,05 mol

=> nHCl pư = 2nMg + 6nAl₂O₃ = 0,55 mol

mHCl = 0,55.36,5 = 20,075 gam

=> mdung dịch HCl 18% = \(\dfrac{20,075}{18\%}\)= 111,53 gam