65) 

Fe + 2HCl ---> FeCl₂ + H₂

2Fe + 3Cl₂ --t^o--> 2FeCl₃

2FeCl₂ + Cl₂ --t^o--> 2FeCl₃

Fe + 2FeCl₃ ---> 3FeCl₂ 

FeCl₂ + 2AgNO₃ ---> Fe(NO₃)₂ + 2AgCl

FeCl₃ + 3AgNO₃ ---> Fe(NO₃)₃ + 3AgCl

6Fe(NO₃)₂ + 3Cl₂ ---> 4Fe(NO₃)₂ + 2FeCl₃

Fe(NO₃)₃ + 3KOH ---> Fe(OH)₂ + 3KNO₃

Fe(NO₃)₂ + 2KOH ---> Fe(OH)₂ + 2KNO₃

4Fe(OH)₂ + O₂ + 2H₂O ---> 4Fe(OH)₃

4Fe(OH)₂ + O₂ --t^o--> 2Fe₂O₃ + 4H₂O

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

66)

a) 

CTHH hoá học của A₁ có dạng \(C_xH_yO_zN_t\left(x,y,z,t\in N\text{*}\right)\)

\(\rightarrow x:y:z:t=\dfrac{3}{12}:\dfrac{1}{1}:\dfrac{4}{16}:\dfrac{7}{14}=1:4:1:2\)

Vậy CTĐGN của A là \(\left(CH_4ON_2\right)\left(n\in N\text{*}\right)\)

Mà trong CTPT của A có 2 nguyên tử N

\(\rightarrow n=\dfrac{2}{2}=1\left(TM\right)\)

Vậy CTPT của A là \(CH_4ON_2\) hay \(\left(NH_2\right)_2CO\)

A₁, A₂, A₃, A₄ lần lượt là (NH₂)₂CO, (NH₄)₂CO₃, CO₂, NH₃

\(2NH_3+CO_2\xrightarrow[200atm]{180-200^oC}\left(NH_2\right)_2CO+H_2O\)

\(\left(NH_2\right)_2CO+2H_2O\rightarrow\left(NH_4\right)_2CO_3\)

\(\left(NH_4\right)_2CO_3+H_2SO_4\rightarrow\left(NH_4\right)_2SO_4+CO_2\uparrow+H_2O\)

\(\left(NH_4\right)_2CO_3+2NaOH\rightarrow Na_2CO_3+2NH_3\uparrow+2H_2O\)

b)

Cách làm khô CO₂: H₂SO_4(đặc)

Cách làm khô NH₃: CaO khan

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,1`   `0,2`                             `0,1`          `(mol)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,1`                       `0,1`                     `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`n_[HCl]=[10,95]/[36,5]=0,3(mol)`

Ta có: `[0,1]/1 < [0,3]/2`

   `=>HCl` dư

`b)V_[H_2]=0,1.22,4=2,24(l)`

   `m_[Cu]=0,1.64=6,4(g)`

A

Khối lượng dd là bao nhiêu?

61)

CaCO₃ --t^o--> CaO + CO₂

CaO + CO₂ ---> CaCO₃

CaO + H₂O ---> Ca(OH)₂

\(CO_2+H_2O⇌H_2CO_3\)

Ca(OH)₂ + H₂CO₃ ---> CaCO₃ + 2H₂O

Ca(OH)₂ + 2HCl ---> CaCl₂ + 2H₂O

2NaOH + CO₂ ---> Na₂CO₃ + H₂O

CaCl₂ + Na₂CO₃ ---> CaCO₃ + 2NaCl

a) Gọi \(n_{Cu}=a\left(mol\right)\rightarrow n_{Fe}=\dfrac{3}{2}a=1,5a\left(mol\right)\)

PTHH: 

\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

a<------a<------a

\(Fe_3O_4+4H_2\xrightarrow[]{t^o}3Fe+4H_2O\)

0,5a<-----2a<------1,5a

\(\rightarrow80a+0,5a.232=39,2\\ \Leftrightarrow a=0,2\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_3O_4}=0,5.0,2.232=23,2\left(g\right)\end{matrix}\right.\)

b) \(V_{H_2}=\left(0,2.2+0,2\right).22,4=13,44\left(l\right)\)

a)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(m_{HCl}=\dfrac{175.14,6}{100}=25,55\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{25,55}{35,5}=0,7\left(mol\right)\)

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

bđ       0,3       0,7

pư      0,3       0,6

spư      0         0,1        0,3       0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b)

m_dd = 16,8 + 175 - 0,3.2 = 191,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,93\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{191,2}.100\%=1,91\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{\dfrac{175.14,6}{100}}{36,5}=0,7\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
\(LTL:\dfrac{0,3}{1}< \dfrac{0,7}{2}\)  
\(n_{H_2}=n_{Fe}=0,3\left(mol\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\\ m_{\text{dd}}=16,8+175-\left(0,3.2\right)=191,2\left(g\right)\\ n_{FeCl_2}=n_{Fe}=0,3\left(mol\right)\\ C\%_{FeCl_2}=\dfrac{0,3.127}{191,2}.100\%=19,92\%\)
=> HCl dư 
 

43)

2KMnO₄ + 16HCl ---> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O
Cl₂ + 2NaOH ---> NaCl + NaClO

NaClO + 2HCl ---> NaCl + Cl₂ + H₂O

2KMnO₄ + 5NaClO₂ + 3H₂SO₄ ---> 5NaClO₃ + 3MnSO₄ + K₂SO₄ + 3H₂O

3Cl₂ + 6NaOH ---> 5NaCl + NaClO₃ + 3H₂O

2NaClO₃ --t^o--> 2NaCl + 3O₂

44)

(1) 4Al + 3O₂ --t^o--> 2Al₂O₃

(2) Al₂O₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂O

(3) Al₂(SO₄)₃ + 3Ba(OH)₂ ---> 2Al(OH)₃ + 3BaSO₄

(4) Al(OH)₃ + NaOH ---> NaAlO₂ + 2H₂O

(5) NaAlO₂ + HCl + H₂O ---> NaCl + Al(OH)₃

(7) 2Al(OH)₃ --t^o--> Al₂O₃ + 3H₂O

(8) Al + Fe(NO₃)₃ ---> Al(NO₃)₃ + Fe

(9) AlCl₃ + 3AgNO₃ ---> Al(NO₃)₃ + 3AgCl

(10) Al(NO₃)₃ + 3KOH ---> Al(OH)₃ + 3KNO₃

45)

\(2H_2O\xrightarrow[]{\text{điện phân}}2H_2+O_2\)

\(H_2+Cl_2\xrightarrow[]{\text{ánh sáng}}2HCl\)

O₂ + 2Ca --t^o--> 2CaO

CaO + 2HCl ---> CaCl₂ + H₂O

4HCl + MnO₂ ---> MnCl₂ + Cl₂ + 2H₂O

CaO + H₂O ---> Ca(OH)₂

Ca(OH)₂ + Cl₂ ---> CaOCl₂ + H₂O

\(n_{N_2}=\dfrac{3,5}{28}=0,125\left(mol\right)\\ n_{O_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ pthh:N_2+O_2\underrightarrow{t^o}2NO\\ LTL:\dfrac{0,125}{1}>\dfrac{0,025}{1}\) 
=> N₂ dư 
\(n_{N_2\left(p\text{ư}\right)}=n_{O_2}=0,025\left(mol\right)\\ m_{N_2\left(d\right)}=\left(0,125-0,025\right).28=2,8\left(g\right)\\ n_{NO}=2n_{O_2}=0,5\left(mol\right)\\ m_{NO}=0,5.30=15\left(g\right)\\ m_{sp}=2,8+15=17,8\left(g\right)\)

`d_[B//CH_4]=[\overline{M_B}]/16=2,5`

   `=>\overline{M_B}=40`

  `=>[n_[N_2 O]]/[n_[O_2]]=|[40-32]/[40-44]|=2`

Vì phần trăm của thể tích mỗi khí trog hỗn hợp `B` bằng phần trăm của số mol mỗi khí trong hỗn hợp `B`

  `=>%V_[N_2 O]=2/[2+1] .100~~66,67%`

  `=>%V_[O_2]~~100-66,67~~33,33%`

\(m_B=\dfrac{0,894}{8,132\%}=11\left(g\right)\)

\(n_C=\dfrac{0,528}{12}=0,44\left(mol\right)\)

Đặt \(n_{O_2}=a\left(mol\right)\)

\(\rightarrow n_{kk}=3a\left(mol\right)\rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{20}{100}.3a=0,6a\left(mol\right)\\n_{N_2}=3a-0,6a=2,4a\left(mol\right)\end{matrix}\right.\)

Vậy khí X gồm: \(\left\{{}\begin{matrix}O_2:n_{O_2}=a+0,6a=1,6a\left(mol\right)\\N_2:n_{N_2}=2,4a\left(mol\right)\end{matrix}\right.\)

+) TH₁: Khí Y gồm CO₂, N₂, O_2(dư)

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

     0,044-->0,044->0,044

\(\rightarrow\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=1,6x-0,044\left(mol\right)\\n_Y=\dfrac{0,044}{22,92\%}=0,2\left(mol\right)\end{matrix}\right.\)

Theo bài ra, ta có PT:

\(2,4a+0,044+16,a-0,044a=0,2\\ \Leftrightarrow a=0,05\left(mol\right)\)

Áp dụng ĐLBTKL:

\(m_A=m_B+m_{O_2}=11+0,05.32=12,6\left(g\right)\)

+) TH₂: khí Y gồm CO₂, N₂, CO

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\) 

        1,6a<-1,6a-->1,6a

\(\rightarrow n_{C\left(dư\right)}=0,044-1,6a\left(mol\right)\)

PTHH: \(C+CO_2\xrightarrow[]{t^o}2CO\)

(0,044-1,6a)->(0,044-1,6a)->(0,088-3,2a)

\(\rightarrow n_{CO_2}=1,6a-\left(0,044-1,6a\right)=3,2a-0,044\left(mol\right)\)

\(\rightarrow n_Y=2,4a+3,2a-0,044+0,088-3,2a=2,4a+0,044\)

Theo bài ra, ta có PT:

\(\dfrac{3,2a-0,044}{2,4a+0,044}.100\%=22,92\%\\ \Leftrightarrow a=0,02\left(mol\right)\)

Áp dụng ĐLBTKL:

\(m_A=m_B+m_{O_2}=11+0,02.32=11,64\left(g\right)\)