a,Fe     +        2HCl            →            FeCl               +              H_{2           (1)}

   FeO   +        2HCl            →            FeCl               +              H₂O       (2)

n_H2 =  3,36/ 22,4 = 0,15 ( mol)

Theo (1)  n_H2 = nFe =  0,15 ( mol)

m_Fe = 0,15 x 56  =  8.4 (g)

m _FeO = 12 - 8,4  =  3,6 (g)

a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)  

\(Fe+2HCl->FeCl_2+H_2\left(1\right)\) 

\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\) 

theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\) 

=> \(m_{Fe}=0,15.56=8,4\left(g\right)\) 

=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)

ta thấy : nFe =nH2 = 0,15

=> mFe =0,15 x 56 = 8,4g

%Fe=8,4/12 x 100 = 70%

=>%FeO = 100 - 70 = 30%

b) BTKLra mdd tìm mct of HCl

c) tìm mdd sau pứ -mH2 nha bạn

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,3\left(mol\right)=n_{ZnCl_2}\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0.3\cdot65}{35,5}\cdot100\%\approx54,93\%\\\%m_{Cu}=45,07\%\\C\%_{HCl}=\dfrac{0,6\cdot36,5}{500}\cdot100\%=4,38\%\\m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{Cu}=35,5-0,3\cdot65=16\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{Cu}-m_{H_2}=518,9\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{518,9}\cdot100\%\approx7,86\%\)

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

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$

Đáp án cần chọn là: A