Bài 30: Ankađien

\(n_{C_4H_6} = n.n_{cao\ su} = n.\dfrac{1000}{54n} = \dfrac{500}{27}(kmol)\\ n_{C_2H_5OH} = 2.\dfrac{1}{60\%}.n_{C_4H_6} = \dfrac{5000}{81}(kmol)\\ n_{C_6H_{12}O_6} = \dfrac{1}{2}.\dfrac{1}{80\%}.n_{C_2H_5OH} = \dfrac{3125}{81}(kmol)\\ m_{gỗ} = \dfrac{m_{C_6H_{12}O_6}}{35\%} = \dfrac{\dfrac{3125}{81}.180}{35\%} = 19841,25(kg)\)

\(CH_2=CH-CH=CH_2 + 2Br_2 \to CH_2Br-CHBr-CHBr-CH_2Br\\ n_{Br_2} = 2n_{buta-1,3-dien} = 2.\dfrac{5,4}{54} = 0,2(mol)\\ \Rightarrow m_{Br_2} = 0,2.160 = 32\ gam\)

\(CH_2=CH-CH_2-CH=CH_2\)

\(CH_2=C=CH-CH_2-CH_3\)

\(CH_2=C-CH=CH-CH_2\)

\(CH_3-CH=CH=CH-CH_3\)

\(CH_2=C\left(CH_3\right)-CH=CH_3\)

\(CH_2-C\left(CH_3\right)=C=CH_3\)

Ankadien liên hợp là : ankadien có 2 liên kết đôi cách nhau bởi 1 liên kết đơn 

=> CT : 3 , 5 

\(CT:C_nH_{2n-2}\)

\(C_nH_{2n-2}+2Br_2\rightarrow C_nH_{2n-2}Br_4\)

\(0.025............0.05\)

\(M_X=\dfrac{5.4}{0.025}=216\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow14n-2=216\)

\(\Rightarrow n=15.5\left(l\right)\)

=> Sai đề

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