Question

cho 30g ch3cooh tác dụng với 80g c2h5oh (h2so4 đặc t độ)

a) pthh

b) tính khối lượng ch3cooc2h5 biết h % = 80%

Answer 1

a, PTHH: CH₃COOH + C₂H₅OH --t^o, H₂SO_4(đ)--> CH₃COOC₂H₅ + H₂O

b, \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\\n_{C_2H_5OH}=\dfrac{80}{46}=\dfrac{40}{23}\left(mol\right)\end{matrix}\right.\)

LTL: \(0,5< \dfrac{40}{23}\) => C₂H₅OH dư

Theo pthh: n_CH3COOC2H5 = n_CH3COOH = 0,5 (mol)

=> \(m_{este}=0,5.80\%.88=35,2\left(g\right)\)

Answer 2

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_4đ,t^o}CH_3COOC_2H_5+H_2O\)

\(nCH_3COOH=\dfrac{30}{60}=0,5\left(mol\right)\)

\(nC_2H_5OH=\dfrac{80}{46}=1,74\left(mol\right)\) 

=> CH3COOH đủ  , nC2H5OH dư

=> \(CH_3COOC_2H_5=0,5\left(mol\right)\)

=> \(mCH_3COOC_2H_5=0,5.88=44\left(g\right)\)

=> \(mCH_3COOC_2H_{5\left(thựcte\right)}=\dfrac{44.80}{100}=35,2\left(g\right)\)