Viết ptpu theo chuỗi sau và ghi các điều kiện phản ứng
a/ butan--> etan--> etyl clorua --> butan --> metan --> cacbon
b/ nhôm cacbua --> metan --> metyl clorua --> etan --> etyl clorua
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a, \(Al_4C_3+12HCl\rightarrow4Al\left(OH\right)_3+3CH_4\)
\(CH_4+Cl_2\underrightarrow{^{as}}CH_3Cl+HCl\)
\(CH_3Cl+Cl_2\rightarrow CH_2Cl_2+HCl\)
\(CH_2Cl_2+Cl_2\rightarrow CHCl_3+HCl\)
\(CHCl_3+Cl_2\rightarrow CCl_4+HCl\)
b,\(C_4H_{10}\rightarrow C_2H_6+C_2H_4\)
\(C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl\)
\(2C_2H_5Cl+2Na\rightarrow C_4H_{10}+2NaCl\)
\(C_4H_{10}\rightarrow C_3H_6+CH_4\)
\(C_3H_8\rightarrow CH_4+C_2H_4\)
\(CH_4+2O_2\rightarrow2H_2O+CO_2\)
c,\(CH_3COONa+NaOH\rightarrow CH_4+Na_2CO_3\)
\(2CH_4\rightarrow C_2H_2+3H_2\)
\(Al_4C_3+12HCl\rightarrow4AlCl_3+3CH_4\)
\(CH_4+2F_2\rightarrow4HF+C\)
\(C+2H_2\rightarrow CH_4\)
\(CH_4+3Cl_2\rightarrow CHCl_3+3HCl\)
\(C_3H_8\rightarrow C_2H_4+CH_4\)
\(CH_4+O_2\rightarrow HCHO+H_2O\)
a,
\(CH_3COONa+NaOH\underrightarrow{^{CaO,to}}CH_4+Na_2CO_3\)
\(2CH_4\underrightarrow{^{1500oC}}C_2H_2+3H_2\)
\(2C_2H_2\underrightarrow{^{to,xt}}C_6H_6\)
b,
\(C_4H_{10}\underrightarrow{^{to,xt}}C_2H_6+C_2H_4\)
\(C_2H_6+Cl_2\rightarrow C_3H_5OH+HCl\)
\(C_2H_5Cl+NaOH\rightarrow C_2H_5OH+NaCl\)
\(C_2H_5OH\underrightarrow{^{to}}C_2H_4+H_2O\)
\(nC_2H_4\underrightarrow{^{to,xt}}\left(C_2H_4\right)_n\)
\(C_2H_6\underrightarrow{^{to,xt}}C_2H_4+H_2\)
\(C_2H_4+H_2O\underrightarrow{^{H^+}}C_2H_5OH\)
\(C_2H_5OH+3O_2\rightarrow2CO_2+3H_2O\)
c,
\(CH_3COONa+NaOH\underrightarrow{^{CaO,to}}CH_4+Na_2CO_3\)
\(2CH_4\underrightarrow{^{1500oC}}C_2H_2+3H_2\)
\(C_2H_2+H_2\underrightarrow{^{Pd,to}}C_2H_4\)
\(C_2H_4+H_2O\underrightarrow{^{H^+}}C_2H_5OH\)
a) CH3COONa + NaOH\(\xrightarrow[t]{CaO}\) CH4 + Na2CO3
CH4+ Cl2 -> CHCl3 +HCl
b) C4H10 \(\xrightarrow[xt]{t}C_3H_6+CH_4\)
CH4 + O2-> CO2 +H2O
CO2 +Ca(OH)2-> CaCO3 +H2O
C) Al4C3 + H2O -> Al(OH)3 + CH4
CH4 \(\xrightarrow[lamlanhnhanh]{1500^0}\) C2H2 +H2
p/s: làm tới đây thôi ,tự cân bằng
a)
\(4Al+3C\xrightarrow[]{1500-1700^oC}Al_4C_3\\ Al_4C_3+12H_2O\rightarrow4Al\left(OH\right)_3\downarrow+3CH_4\uparrow\\ CH_4+Cl_2\xrightarrow[]{a/s}CH_3Cl+HCl\\ CH_3Cl+Cl_2\xrightarrow[]{a/s}CH_2Cl_2+HCl\\ CH_2Cl_2+Cl_2\xrightarrow[]{a/s}CHCl_3\\ CHCl_3+Cl_2\xrightarrow[]{a/s}CCl_4+HCl\)
b)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4\uparrow+Na_2CO_3\)
\(CH_4+Cl_2\xrightarrow[]{a/s}CH_3Cl+HCl\\ 2CH_3Cl+2Na\xrightarrow[]{t^o}2NaCl+C_2H_6\\ C_2H_6\xrightarrow[]{t^o,xt}C_2H_4\)
c)
\(C_4H_{10}\xrightarrow[]{crackinh}C_2H_6+C_2H_4\\ C_2H_6+Cl_2\xrightarrow[]{a/s}C_2H_5Cl+HCl\\ 2C_2H_5Cl+2Na\xrightarrow[]{t^o,xt}C_4H_{10}+2NaCl\\ C_4H_{10}\xrightarrow[]{crackinh}CH_4+C_3H_6\\ C_3H_6+H_2\xrightarrow[]{Ni,t^o}C_3H_8\)
\(a,\)
\(CH_2COOHNa+NaOH\underrightarrow{^{to,CaO}}CH_4+Na_2CO_3\)
\(CH_4+Cl_2\underrightarrow{^{askt}}CH_3Cl+HCl\)
\(CH_3Cl+Cl_2\underrightarrow{^{to}}CH_2Cl_2+HCl\)
\(CH_2Cl_2+Cl_2\underrightarrow{^{askt}}CHCL_3+HCl\)
\(CH_3Cl_2+Cl_2\underrightarrow{^{askt}}CCl_4+HCl\)
\(b,\)
\(Al_4Cl_3+12HCl\rightarrow4AlCl_3+3CH_4\)
\(CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\)
+\(CH_4\) \(\underrightarrow{1500^oC}\) \(C_2H_2+H_2\)
\(C_2H_2+H_2\) \(\underrightarrow{Pd\text{/}PbCO_3,t^o}\) \(C_2H_4\)
\(C_2H_4\)\(+H_2O\) \(\underrightarrow{xt,t^o}\) \(C_2H_5OH\)
\(2C_2H_5OH+O_2\) \(\underrightarrow{men,t^o}\) \(2CH_3COOH\)
\(C_2H_5OH\) \(+CH_3COOH\) \(\underrightarrow{H_2SO_4,t^o}\) \(CH_3COOC_2H_5+H_2O\)
+\(C_2H_2+HCl\) \(\underrightarrow{xt,t^o}\) \(CH_2=CHCl\)
\(nCH_2=CHCl\) \(\underrightarrow{xt,p,t^o}\) \(\left(-CH_2-CHCl-\right)_n\)
Đáp án A
Các khi đo ở cùng điều kiện nên
xem thể tích như số mol.
C4H10 → c r a c k i n h anken + ankan
10 0 0
x x x
10-x x x
→ nsau phản ứng = 10 + x = 18 → x = 8 → H = 80%.
a,
\(C_4H_{10}\underrightarrow{^{to}}C_2H_4+C_2H_6\)
\(C_2H_6+Cl_2\underrightarrow{^{als}}C_2H_5Cl+HCl\)
\(2C_2H_5Cl+Na\underrightarrow{^{to}}C_4H_{10}+2NaCl\)
\(C_4H_{10}\underrightarrow{^{to}}CH_4+C_3H_6\)
\(CH_4+2Cl_2\rightarrow C+4HCl\)
b,
\(Al_4C_3+12HCl\rightarrow3CH_4+4AlCl_3\)
\(CH_4+Cl_2\underrightarrow{^{as}}CH_3Cl+HCl\)
\(2CH_3Cl+Na\underrightarrow{^{to}}C_2H_6+2NaCl\)
\(C_2H_6+Cl_2\underrightarrow{^{as}}C_2H_5Cl+HCl\)
a) - C4H10 -to-> C2H6 + C2H4
- C2H6 + Cl2 --> C2H5Cl + HCl
- 2C2H5Cl + Na --> C4H10 + 2NaCl
- C4H10 -to-> CH4 + C3H6
- CH4 --to--> C + 2H2
b) - Al4C3 + 12H2O --> 4Al(OH)3 + 3CH4
- CH4 + Cl2 --> CH3Cl + HCl
- 2CH3Cl + Na -to-> C2H6 + 2NaCl
- C2H6 + Cl2 --> C2H5Cl + HCl