Fe + 2HCl -> FeCl2 + H2

          0.2                     0.1

FeO + 2HCl -> FeCl2 + H2O

 0.1       0.2

a.\(nH2=\dfrac{2.24}{22.4}=0.1mol\)

\(\%mFe=\dfrac{0.1\times56\times100}{12.8}=43.8\%\)

\(\%mFeO=100-43.8=56.2\%\)

b.\(nFeO=\dfrac{12.8-\left(0.1\times56\right)}{56+16}=0.1mol\)

\(V_{HCl}=\dfrac{0.2+0.2}{2}=0.2l\)

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

a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,1     0,2         0,1            0,1

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:     0,05         0,3          0,1     

\(m_{Fe}=0,1.56=5,6\left(g\right);m_{Fe_2O_3}=13,6-5,6=8\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

b,\(m_{HCl}=\left(0,2+0,3\right).98=49\left(g\right)\)

\(m_{ddHCl}=\dfrac{49.100}{10}=490\left(g\right)\)

c,\(m_{ddX}=13,6+490-0,1.2=503,4\left(g\right)\)

d,\(C\%_{FeCl_2}=\dfrac{0,1.127.100\%}{503,4}=2,52\%\)

  \(C\%_{FeCl_3}=\dfrac{0,05.162,5.100\%}{503,4}=1,614\%\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow n_{Fe}=n_{H_2}=n_{FeCl_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\approx41,176\%\\ \Rightarrow\%m_{Fe_2O_3}\approx58,824\%\\ b.n_{Fe_2O_3}=\dfrac{13,6-0,1.56}{160}=0,05\left(mol\right)\\ n_{HCl}=2.n_{Fe}+6.n_{Fe_2O_3}=2.0,1+6.0,05=0,5\left(mol\right)\\ m_{ddHCl}=\dfrac{0,5.36,5.100}{10}=182,5\left(g\right)\)

\(c.m_{ddX}=m_{hh\left(Fe,Fe_2O_3\right)}+m_{ddHCl}-m_{H_2}=13,6+182,5-0,1.2=195,9\left(g\right)\\ d.n_{FeCl_3}=2.0,05=0,1\left(mol\right)\\ C\%_{ddFeCl_3}=\dfrac{0,1.162,5}{195,9}.100\approx8,295\%\\ C\%_{ddFeCl_2}=\dfrac{0,1.127}{195,9}.100\approx6,483\%\)

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\) 

        1        2               1           1

       0,1     0,2            0,1         0,1

     \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)

         1            6              2             3

      0,05        0,3            0,1  

a) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Fe}=01,.56=5,6\left(g\right)\)

\(m_{Fe2O3}=13,6-5,6=8\left(g\right)\)

⁰/₀Fe =  \(\dfrac{5,6.100}{13,6}=41,18\)⁰/0

⁰/₀Fe₂O₃ = \(\dfrac{8.100}{13,6}=58,82\)⁰/₀

b) \(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)

⇒ \(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{10}=182,5\left(g\right)\)

c) \(n_{FeCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{FeCl2}=0,1.127=12,7\left(g\right)\)

\(n_{FeCl3}=\dfrac{0,3.2}{6}=0,1\left(mol\right)\)

⇒ \(m_{FeCl3}=0,1.162,5=16,25\left(g\right)\)

d) \(m_{ddspu}=13,6+182,5-\left(0,1.2\right)=195,9\left(g\right)\)

\(C_{FeCl2}=\dfrac{12,7.100}{195,9}=6,48\)⁰/₀

\(C_{FeCl3}=\dfrac{16,25.100}{195,9}=8,3\)⁰/₀

 Chúc bạn học tốt

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)

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

Đáp án : D

+) A + HCl : 2 khí là H₂ và CO₂ => n_H2 = n_CO2 = 0,06 mol

=> n_Fe = n_FeCO3 = 0,06 mol

+) A + HNO₃ -> 1 muối là Fe(NO₃)₃ có n = 0,4 mol

Hỗn hợp khí gồm CO₂ và khí hóa nâu ngoài không khí là NO

=> n_CO2 + n_NO = 0,2 mol => n_NO = 0,14 mol

Bảo toàn e : 3n_Fe + n_{Fe2+(oxit,hidroxit)} + n_FeCO3 = 3n_NO

=> n_{Fe2+(oxit,hidroxit)} = 0,18 mol

Bảo toàn Fe : n_{Fe2+(oxit,hidroxit) +} n_{Fe3+(oxit,hidroxit)} + n_Fe + n_FeCO3 = 0,4 mol

=> n_{Fe3+(oxit,hidroxit)} = 0,1 mol

+) A + HCl -> muối gồm FeCl₂ và FeCl₃

=> n_FeCl3 = n_{Fe3+(oxit,hidroxit)} = 0,1 mol

Và n_FeCl2 = n_{Fe2+(oxit,hidroxit)} + n_Fe + n_FeCO3 = 0,3 mol

=> m = 54,35g

a,Giả sử m_ddHCl = 36,5 (g) \(\Rightarrow n_{HCl}=\dfrac{36,5.0,2}{36,5}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     a         2a          a         a

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:       b         2b           b         b

Ta có: \(2a+2b=0,2\Leftrightarrow a+b=0,1\left(mol\right)\)

m_{dd D} = 56a+24b+36,5-(a+b).2 = 56a+24b+36,3 (g)

\(C\%_{ddFeCl_2}=\dfrac{127a.100\%}{56a+24b+36,3}=15,757\%\)

\(\Leftrightarrow127a=8,82392a+3,78168b+5,719791\)

\(\Leftrightarrow118,17608a-3,78168b=5,719791\)

Ta có: \(\left\{{}\begin{matrix}a+b=0,1\\118,17608a-3,78168b=5,719791\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,05\end{matrix}\right.\)

\(C\%_{ddMgCl_2}=\dfrac{95.0,05.100\%}{56.0,05+24.0,05+36,3}=11,79\%\)