nH2 = 6,72/22,4 = 0,3 (mol)

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

nAl = 0,3 : 3 . 2 = 0,2 (mol)

nHCl (Al) = 0,3 . 2 = 0,6 (mol)

mAl = 0,2 . 27 = 5,4 (g)

%mAl = 5,4/25,65 = 20,05%

%mZnO = 100% - 20,05% = 79,95%

mZnO = 25,65 - 5,4 = 20,25 (g)

nZnO = 20,25/81 = 0,25 (mol)

PTHH: ZnO + 2HCl -> ZnCl2 + H2O

nHCl (ZnO) = 0,25 . 2 = 0,5 (mol)

nHCl (đã dùng) = 0,6 + 0,5 = 1,1 (mol)

CMddHCl = 1,1/0,1008 = 10,9M

C% = (10,9 . 36,5)/(10 . 1,19) = 33,43%

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

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

a, \(n_A=0,5\left(mol\right)=n_{H_2}+n_{CO_2}\)

\(m_{hh}=33=24n_{Mg}+84n_{MgCO_3}=84n_{CO_2}+24n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,15\left(30\%\right)\\n_{CO_2}=0,35\left(70\%\right)\end{matrix}\right.\) ( mol )

b, Có \(n_{HCl}=2n_{H_2}+2n_{CO_2}=1\left(mol\right)\)

Mà \(n_{HClbd}=1,6\left(mol\right)\)

\(\Rightarrow n_{HCldu}=0,6\left(mol\right)\)

Lại có : \(m_{ddsau}=m_{hh}+m_{ddHCl}-m_{hhkhi}=33+880-0,15.2-0,35.44=897,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,5.\left(24+71\right)}{897,3}.100\%\approx5,3\%\\C\%_{HCldu}=\dfrac{0,6\left(35,5+1\right)}{897,3}.100\%\approx2,44\%\end{matrix}\right.\)

\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=5,6\left(g\right)\)

\(\Rightarrow m_{Fe_2O_3}=16\left(g\right)\)

\(\Rightarrow n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)=n_{FeCl_3}\)

Lại có : \(n_{HCl}=2n_{H_2}+3n_{FeCl_3}=0,8\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=292\left(g\right)\)

\(\Rightarrow V=\dfrac{2920}{11}\left(ml\right)=\dfrac{73}{275}\left(l\right)\)

\(\Rightarrow C_{MFeCl_3}=\dfrac{0,2}{\dfrac{73}{275}}=\dfrac{55}{73}\left(M\right)\)

\(n_{H_2} = 1(mol)\)

Fe + 2HCl → FeCl₂ + H₂

1.........2.......................1.............(mol)

⇒ m_Fe = 1.56 = 56 > m_{hỗn hợp} = 21,6

(Sai đề)

a) Bảo toàn nguyên tố H : \(n_{HCl}.1=2n_{H_2}=0,6\left(mol\right)\)

=> nH2=0,3(mol)

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

b) Áp dụng định luật bảo toàn khối lượng :

 \(m_{ct}=m_{kl}+m_{HCl}-m_{H_2}=10,4+0,6.36,5-0,3.2=31,7\left(g\right)\)

Đáp án B.