\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,2------------>0,2----->0,2

=> \(\left\{{}\begin{matrix}m_{FeCl_2}=0,2.127=25,4\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,1______0,2_____0,1____0,1 (mol)

a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

Ở đkc, 1 mol khí chiếm thể tích 24,79 lít

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đkc\right)}=24,79\cdot0,2=4,958\left(l\right)\)

pứ: Fe + 2HCl -> FeCl₂ + H₂

b. n_Fe = \(\dfrac{5,6}{56}\)= 0,1 mol

Từ pt suy ra được: n_HCl = 2.n_Fe= 0,2 mol

=> m_HCl = 0,2. 36,5 = 7,3 g

c. n_H2 = n_Fe = 0,1 mol

=> V_H2 = 0,1.22,4 = 2,24 (lít)

\(1.\\ n_A=\dfrac{16,8}{A}mol\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ A+2HCl\rightarrow ACl_2+H_2\\ \Rightarrow\dfrac{16,8}{A}=0,3\\ \Rightarrow A=56g/mol\\ \Rightarrow A.là.Fe\\ \Rightarrow Chọn.A\\ 2.\\ n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ \Rightarrow Chọn.B\\ 3.\\ Axit:H_2SO_4;HCl\\ \Rightarrow Chọn.B\\ 4.\\ 3,719l\Rightarrow3,7185\\ CTHH:R\\ n_R=\dfrac{3,6}{R}mol\\ n_{H_2}=\dfrac{3,7185}{24,79}=0,15mol\\ R+2HCl\rightarrow RCl_2+H_2\\ \Rightarrow\dfrac{3,6}{R}=0,15\\ \Rightarrow R=24g/mol,Mg\\ \Rightarrow Chọn.B\)

\(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)

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

      0,05<--0,1----->0,05--->0,05

\(m_{Fe}=0,05.56=2,8\left(g\right)\)

\(m_{FeCl2}=0,05.127=6,35\left(g\right)\)

\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

\(V_{H2\left(dkc\right)}=0,05.24,79=1,2395\left(l\right)\)

\(4.\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

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

\(0.15.....0.3....................0.15\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.3}{0.5}=0.6\left(M\right)\)

\(5.\)

\(Đặt:n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_{hh}=56a+27b=8.3\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

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

\(\Rightarrow a+1.5b=0.25\left(2\right)\)

\(\left(1\right),\left(2\right):a=b=0.1\)

\(\%Fe=\dfrac{5.6}{8.3}\cdot100\%=67.47\%\)

\(\%Al=32.53\%\)

9.A

10.B

11.C

12.C

13.D

a) PT: Fe+2HCl→FeCl₂+H₂   (1) 

- Số mol Fe là:

n_Fe=\(\dfrac{m}{M}\)=\(\dfrac{11,2}{56}\)=0,2(mol)

- Theo PT (1)⇒n_FeCl2=n_Fe=0,2(mol)

- Vậy khối lượng của FeCl₂ là:

m_FeCl2=n.M=0,2.127=25,4(g)

b) Theo PT (1)⇒n_H2=n_Fe=0,2(mol)

- Vậy thể tích của H₂ là:

V_H2=n.24,79=0,2.24,79=4,958(l)

`#3107.101107`

`a)`

\(\text{Fe + 2HCl}\rightarrow\text{FeCl}_2+\text{H}_2\)

n của Fe có trong phản ứng là:

\(\text{n}_{\text{Fe}}=\dfrac{\text{m}_{\text{Fe}}}{\text{M}_{\text{Fe}}}=\dfrac{11,2}{56}=0,2\left(\text{mol}\right)\)

Theo PT: \(\text{n}_{\text{Fe}}=\text{n}_{\text{ }\text{FeCl}_2}=0,2\left(\text{mol}\right)\)

m của FeCl₂ có trong phản ứng là:

\(\text{m}_{\text{FeCl}_2}=\text{n}_{\text{FeCl}_2}\cdot\text{M}_{\text{FeCl}_2}=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(\text{g}\right)\)

`b)`

Theo PT: \(\text{n}_{\text{Fe}}=\text{n}_{\text{H}_2}=0,2\left(\text{mol}\right)\)

V của khí H₂ ở đkc là:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,2\cdot24,79=4,958\left(\text{l}\right)\)`.`