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)

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

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

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

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

=> CH3COOH đủ  , nC2H5OH dư

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

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

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

a) 

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                0,1--------->0,2------------->0,2------------>0,1

=> m_CH3COOH = 0,2.60 = 12 (g)

\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

                 0,16------------------------------------->0,16

=> \(m_{CH_3COOC_2H_5}=0,16.88=14,08\left(g\right)\)

CH3COOH+C2H5OH->CH3COOC2H5+H2O

0,5--------------------------------------------0,5

n CH3COOH=0,5 mol

=>m CH3COOC2H5=0,5.88=44g

H=66%

=>V CH3COOH=\(\dfrac{44.66\%}{0,897}\)=32,37l

Chọn đáp án B

n C 2 H 5 O H = 0 , 1   m o l ; n C H 3 C O O H = 0 , 1   m o l

C H 3 C O O H + C 2 H 5 O H → C H 3 C O O C 2 H 5 + H 2 O

n C H 3 C O O C 2 H 5 = 0 , 1 × 88 = 8 , 8   g a m

Đáp án B

n_C2H5OH = 0,1 mol; n_CH3COOH = 0,1 mol
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O
n_CH3COOC2H5 = 0,1 × 88 = 8,8 gam.

a) nCH3COOH= 0,4(mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

0,4____________0,4(mol)

=> mNaOH=0,4. 40=16(g)

b) nCH3COOH= 1(mol)

nC2H5OH= 100/46= 50/23(mol)

Vì : 1/1< 50/23 :1

=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.

PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H⁺ , nhiệt độ)

Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)

Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)

=>H= (0,625/1).100=62,5%

Đáp án A

nC2H5OH = 0,5; nCH3COOH = 0,4

  mCH3COOC2H5 = 0,4.88.60% = 21,12g

  Chọn C.

Đáp án C

nC₂H₅OH = 0,5; nCH₃COOH = 0,4 => mCH₃COOC₂H₅ = 0,4.88.60% = 21,12g