Bài 38. Axetilen

`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, \(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

b, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c, \(n_{\left(CH_3COO\right)_2Cu}=n_{Cu}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Cu}=0,1.182=18,2\left(g\right)\)

Bài 7:

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\a, PTHH:Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ b,n_{H_2}=n_{Zn}=0,4\left(mol\right)\\ V_{B\left(đktc\right)}=V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ c,V_{B\left(có.hiệu.suất\right)}=8,96.90\%=8,064\left(l\right)\)

Bài 8:

Hoà tan 8g gì em?

Ta có: \(n_{C_2H_2}=\dfrac{168}{22,4}=7,5\left(mol\right)\)

PT: \(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\)

Theo PT: \(n_{CaC_2\left(LT\right)}=n_{C_2H_2}=7,5\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{CaC_2\left(TT\right)}=\dfrac{7,5}{90\%}=\dfrac{25}{3}\left(mol\right)\)

\(\Rightarrow m_{CaC_2\left(TT\right)}=\dfrac{25}{3}.64=\dfrac{1600}{3}\left(g\right)\)

Mà: Đất đèn chứa 80% CaC₂.

⇒ m _{đất đèn} = 1600/3 : 80% = 2000/3 (g)

a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

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

\(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,025\left(mol\right)\Rightarrow m_{C_2H_2}=0,025.26=0,65\left(g\right)\)

b, \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,025.22,4}{33,6}.100\%\approx1,67\%\\\%V_{CH_4}\approx98,33\%\end{matrix}\right.\)

a) \(n_{C_2H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)

             0,6----->1,5

b) \(V_{kk}=\dfrac{1,5.22,4}{20\%}=168\left(l\right)\)

Câu 1:

a) \(n_{C_2H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: \(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)

           0,05--->0,125

\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)

b) \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

      0,05-->0,1

\(\Rightarrow V=V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)

Câu 2:

a) \(n_{C_2H_2}=\dfrac{16,8}{26}=\dfrac{42}{65}\left(mol\right)\)

PTHH: \(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)

              \(\dfrac{42}{65}\)----->\(\dfrac{21}{13}\)----->\(\dfrac{84}{65}\)

\(\Rightarrow V_{kk}=5.\dfrac{21}{13}.22,4=\dfrac{2352}{13}\left(l\right)\)

b) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                         \(\dfrac{84}{65}\)----->\(\dfrac{84}{65}\)

\(\Rightarrow m_{CaCO_3}=m_{kt}=\dfrac{84}{65}.100=\dfrac{1680}{13}\left(g\right)\)

a) \(n_{C_2H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

                0,2--->0,5

=> V_O2 = 0,5.22,4 = 11,2 (l)

b) V_kk = 11,2 : 20% = 56 (l)

Số mol C₂H₄ = 4,48 : 22,4 = 0,2 mol

Phương trình phản ứng đốt cháy etilen:

               C₂H₄ + 3O₂  t0→→t0 2CO₂ + 2H₂O

Pư:          0,2      0,6        0,4          (mol)

a) VO2VO2 =0,6 x 22,4 = 13,44 lít

b)