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a)
C2H4 + H2O \(\xrightarrow{t^o,xt}\) C2H5OH
C2H5OH + O2 \(\xrightarrow{men\ giấm}\) CH3COOH + H2O
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
CH3COOC2H5 + KOH → CH3COOK + C2H5OH
b)
(1) 2C2H5OH + 2Na → 2C2H5ONa + H2
(2) 2CH3COOH + Mg → (CH3COO)2Mg + H2
(3) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
(4) (RCOO)3C3H5 + 3H2O ⇌ 3RCOOH + C3H5(OH)3
\(a) C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ b) n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{9,2}{46} = 0,2(mol)\\ m_{CH_3COOH} = 0,2.60 = 12(gam)\\ c) n_{CH_3COOC_2H_5} = n_{C_2H_5OH} = 0,2(mol)\\ m_{CH_3COOC_2H_5} = 0,2.88 = 17,6(gam)\)
a, PT: \(CH_3COOH+C_2H_5OH\underrightarrow{_{H_2SO_{4\left(đ\right)}}}CH_3COOC_2H_5+H_2O\)
b, Ta có: \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{27,6}{46}=0,6\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\), ta được C2H5OH.
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\)
Mà: m CH3COOC2H5 (TT) = 35,2 (g)
\(\Rightarrow H\%=\dfrac{35,2}{44}.100\%=80\%\)
Bạn tham khảo nhé!
a.
\(C_2H_2+H_2\xrightarrow[Pd]{t^o}C_2H_4\)
\(C_2H_4+H_2\xrightarrow[Ni]{t^o}C_2H_6\)
b.
\(C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_4đặc]{t^o}CH_3COOC_2H_5+H_2O\)