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khoi luong mol trung binh cua 2 anken la 1,35.28=37,8$\rightarrow n=2,7\Rightarrow2 anken la C2H4 va C3H6,tu KLM=37,8\Rightarrow$ ti le mol C2H2 :C3H6=3:7(so do duong cheo) goi mol C2H2=3x,mol C3H6=7x, ancol bac 2 la CH3CH(OH)CH3 co m=50$\Rightarrow mol=\frac{5}{6}\Rightarrow$ mol CH3CH2CH2OH=7x-$\frac{5}{6}\Rightarrow$ KL cua CH3CH2OH v CH3CH2CH2OH=43 tu do biet x
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