a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(\left(1\right)\Rightarrow n_{HCl}=2.0,2=0,4\left(mol\right)\)

\(m_{dd}\left(HCl\right)=\dfrac{0,4.36,5}{20}.100=73\left(g\right)\)

b) \(\left(1\right)\Rightarrow n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(m_{H_2}=0,2.2=0,4\left(g\right)\)

\(m_{dd}\left(sau.pư\right)=m_{Zn}+m_{dd}\left(HCl\right)-m_{H_2}=13+73-0,4=85,6\left(g\right)\)

\(C\%\left(ZnCl_2\right)=\dfrac{27,2}{85,6}.100\%=31,78\%\)

`a) Fe + 2HCl -> FeCl_2 + H_2`

`b) n_{Fe} = (5,6)/(56) = 0,1 (mol)`

Theo PT: `n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,1 (mol)`

`Q = 0,1.24,79 = 2,479(l)`

`c)` Theo PT: `n_{HCl} = 2n_{Fe} = 0,2 (mol)`

`m = (0,2.36,5)/(14,6\%) = 50(g)`

`d) m_{dd} = 50 + 5,6 - 0,1.2 = 55,4(g)`

`C\%_{FeCl_2} = (0,1.127)/(55,4) .100\% = 22,92\%`

`a)Zn + H_2SO_4 -> ZnSO_4 + H_2`

`b) n_{Zn} = (13)/(65) = 0,2 (mol)`

Theo PT: `n_{H_2SO_4} = n_{ZnSO_4} = n_{H_2} = n_{Zn} = 0,2 (mol)`

`Q = 0,2.24,79 = 4,958(l)`

`c)m = (0,2.98)/(9,8\%) = 200(g)`

`d) m_{dd} = 13 + 200 - 0,2.2 = 212,6(g)`

`C\%_{ZnSO_4} = (0,2.161)/(212,6) .100\% = 15,15\%`

Ta có: \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a. PTHH: 

Na₂SO₃ + H₂SO₄ ---> Na₂SO₄ + SO₂ + H₂O (1)

K₂SO₃ + H₂SO₄ ---> K₂SO₄ + SO₂ + H₂O (2)

Theo PT_(1,2): \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

b. Gọi x, y lần lượt là số mol của Na₂SO₃ và K₂SO₃

Theo PT_(1,2): 

=> x + y = 0,3

Theo đề, ta có: 126x + 158y = 44,2

=> x = 0,1, y = 0,2

Theo PT₍₁₎: \(n_{Na_2SO_4}=n_{Na_2SO_3}=0,1\left(mol\right)\)

=> \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

Theo PT₍₂₎: \(n_{K_2SO_4}=n_{K_2SO_3}=0,2\left(mol\right)\)

=> \(m_{K_2SO_4}=0,2.164=34,8\left(g\right)\)

=> \(m_{sau.phản.ứng}=34,8+14,2=49\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{0,3.98}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

Ta có: \(m_{dd_{sau.phản.ứng}}=147+44,2=191,2\left(g\right)\)

=> \(C_{\%_{dd_{sau.phản.úng}}}=\dfrac{49}{191,2}.100\%=25,63\%\)

c. Đổi 500ml = 0,5 lít

=> \(n_{Ba\left(OH\right)_2}=1.0,5=0,5\left(mol\right)\)

Ta có: \(T=\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=\dfrac{0,3}{0,5}=0,6< 1\)

Vậy PTHH là: \(SO_2+Ba\left(OH\right)_2--->BaSO_3+H_2O\) (Ba(OH)₂ dư.)

Theo PT: \(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\)

=> \(m_{BaSO_3}=0,3.217=65,1\left(g\right)\)

\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

 \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\)

     \(x\)                                                             \(x\)  

  \(K_2SO_3+H_2SO_4\rightarrow K_2SO_3+H_2O+SO_2\)

    \(y\)                                                            \(y\)

Ta có: \(\left\{{}\begin{matrix}126x+158y=44,2\\x+y=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

a)\(\Rightarrow\Sigma n_{H_2SO_4}=0,1+0,2=0,3mol\)

    \(\Rightarrow m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\)

    \(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{20}\cdot100=147\left(g\right)\)

b) 

`a) n_{khí} = (4,958)/(24,79) = 0,2 (mol)`

`C_2H_4 + Br_2 -> C_2H_4Br_2`

`n_{C_2H_4} = (1,4)/(28) = 0,05(mol)`

`V_{C_2H_4} = 0,05.24,79 = 1,2395(l)`

`V_{CH_4} = 4,958 - 1,2395 = 3,7185(l)`

`b) n_{Br_2} = n_{C_2H_4} = 0,05(mol)`

`=> V_{ddBr_2} = (0,05)/2 = 0,025(l) = 25(ml)`

`c)`

Dẫn qua dd `Br_2` dư, sau đó dẫn qua dd `Ca(OH)_2` dư rồi dẫn qua `CaO` để làm khô thì thu được `CH_4` tinh khiết

`C_2H_4 + Br_2 -> C_2H_4Br_2`

`CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O`

`CaO + H_2O -> Ca(OH)_2`

a, Cho e: \(\overset{0}{Mg}\rightarrow\overset{2+}{Mg}+2e\)

Nhận e: \(\overset{0}{Cl_2}+2e\rightarrow2Cl^-\)

b, Liên kết ion.

\(Na+Cl\rightarrow NaCl\)

Liên kết trong phân tử NaCl thuộc loại liên kết ion

Bài giải

\(CTPT\left(A\right):C_xH_y\)

Theo đè bài ta có :

\(n\left(C\right):n\left(H\right)=\dfrac{81,81}{12}:\dfrac{18,18}{1}=6,82:18,18=1:2,66=3:8=x:y\)

\(\Rightarrow CTPT\left(A\right):\left(C_3H_8\right)_n\)

\(d\left(A/H_2\right)=22\Rightarrow M\left(A\right)=22.2=44\)

\(\Rightarrow44n=44\Rightarrow n=1\)

Vậy \(CTPT\left(A\right):C_3H_8\)

\(CTCT\left(A\right):\) 

a) Chỉ có propylene \(C_3H_6\) phản ứng cộng với brom, ethane \(C_2H_6\) không phản ứng.

\(C_3H_6+Br_2\rightarrow C_3H_6Br_2\left(1\right)\)

\(n\left(Br_2\right)=\dfrac{8}{160}=0,05\left(mol\right)\)

\(\left(1\right)\Rightarrow n\left(C_3H_6\right)=0,05\left(mol\right)\)

\(n\left(hỗn.hợp\right)=\dfrac{V_{hh}}{22,4}=\dfrac{12,395}{22,4}=0,55\left(mol\right)\)

\(\Rightarrow n\left(hỗn.hợp\right)=n\left(C_2H_6\right)+0,05=0,55\)

\(\Rightarrow n\left(C_2H_6\right)=0,55-0,05=0,5\left(mol\right)\)

\(\%V\left(C_2H_6\right)=\dfrac{0,5}{0,55}.100\%=90,91\%\)

\(\Rightarrow\%V\left(C_3H_6\right)=100\%-90,91\%=9,09\%\)

\(m\left(C_2H_6\right)=0,5.30=15\left(g\right)\)

\(m\left(C_3H_6\right)=0,05.42=2,1\left(g\right)\)

\(\%m\left(C_2H_6\right)=\dfrac{15}{15+2,1}.100\%=87,73\%\)

\(\%m\left(C_3H_6\right)=100\%-87,73\%=12,27\%\)

b) \(C_2H_6+\dfrac{7}{2}O_2\rightarrow2CO_2+3H_2O\left(2\right)\)

\(C_3H_6+\dfrac{9}{2}O_2\rightarrow3CO_2+3H_2O\left(3\right)\)

\(\left(2\right);\left(3\right)\Rightarrow n\left(O_2\right)=0,5.3,5+0,05.4,5=1,9\left(mol\right)\)

\(V\left(O_2\right)=1,9.22,4=42,56\left(lít\right)\)

\(V\left(kk\right)=\dfrac{42,56}{20\%}=212,8\left(lít\right)\)

\(\left(2\right);\left(3\right)\Rightarrow n\left(CO_2\right)=0,5.2+0,05.3=1,15\left(mol\right)\)

\(V\left(CO_2\right)=1,15.22,4=25,76\left(lít\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)