\(nCH_2=CH_2\rightarrow\left(-CH_2-CH_2-\right)_n\)

\(CH_4\underrightarrow{^{H2,1400oC}}C_2H_2\rightarrow H_2C=CH-C\equiv CH\underrightarrow{^{+H2}}H_2C=\)

\(CH-CH=CH_2\underrightarrow{^{to,xt}}-\left(-CH_2-CH=CH-CH_2-\right)_n-\)

Chọn B

Chọn đáp án B

Để ý thấy các chất trong X là

đều có 4 nguyên tử C

Đáp án B

1. \(2CH_4\underrightarrow{^{1500^oC,lln}}C_2H_2+3H_2\)

 \(2C_2H_2\underrightarrow{t^o,p,xt}C_4H_4\)

\(CH_2=CH-C\equiv CH+H_2\underrightarrow{t^o,Pd}CH_2=CH-CH=CH_2\)

\(nCH_2=CH-CH=CH_2\underrightarrow{t^o,xt,p}\left(-CH_2-CH=CH-CH_2-\right)_n\)

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

\(CH\equiv CH+HCl\rightarrow CH_2=CHCl\)

\(nCH_2=CHCl\underrightarrow{t^o,p,xt}\left(-CH_2-CHCl-\right)_n\)

1. \(2CH_4\xrightarrow[1500^o]{làm.lạnh.nhanh}CH\equiv CH+3H_2\)

\(CH\equiv CH\xrightarrow[CuCl,NH_4Cl]{t^o}CH_2=CH-C\equiv CH\)

\(CH_2=CH-C\equiv CH+H_2\xrightarrow[Pd,PbCO_3]{t^o}CH_2=CH-CH=CH_2\)

\(nCH_2=CH-CH=CH_2\underrightarrow{t^o,xt,p}\left(-CH_2-CH=CH-CH_2-\right)n\)

2. \(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+CH\equiv CH\)

\(CH\equiv CH+HCl\xrightarrow[HgCl_2]{t^o}CH_2=CHCl\)

\(nCH_2=CHCl\underrightarrow{t^o,xt,p}CH_2-CHCl\)

Chọn đáp án D

n c a o s u = 540 × 10 3 54 n = 10 4 n ⇒ n C 2 H 5 O H 1 t = 10 4 n × n × 2 = 2 × 10 4 ⇒ n C 2 H 5 O H t t = 2 × 10 4 0 ٫ 8 = 25000 ⇒ m C 2 H 5 O H = 25000 × 46 = 1150 × 10 3 g = 1150 k g