a)

$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

b)

n CH3COOC2H5 = n C2H5OH = 9,2/46 = 0,2(mol)

=> m este = 0,2.88 = 17,6 gam

c)

n este = 8,8/88 = 0,1(mol)

=> n C2H5OH = n CH3COOH = 0,1/60% = 1/6 mol

=> m C2H5OH = 46 . 1/6 = 7,67(gam) ; m CH3COOH = 60 . 1/6 = 10(gam)

Đáp án A

\(PTHH:\)

\(CH_3COOH+C_2H_5OH<=(H_2SO_4đ,t^o)=>CH_3COOC_2H_5+H_2O\)

\(n_{CH_3COOH}=\dfrac{80}{60}=\dfrac{4}{3}=1,33\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\)

=> Chọn số mol cảu C2H5OH để tính

Theo PTHH: \(n_{CH_3COOC_2H_5}\left(lt\right)=0,5\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5}\left(lt\right)=0,5.88=44\left(g\right)\)

Hiệu suất phản ứng este hóa

\(H=\dfrac{33}{44}.100\%=75\%\)

nC2H5OH = 8.05/46 = 0.175 (mol) 

nCH3COOH = 36/60 = 0.6 (mol) 

nCH3COOC2H5 = 12.32/88 = 0.14 (mol) 

C2H5OH + CH3COOH <-H2SO4đ,t⁰-> CH3COOC2H5 + H2O 

1.......................1

0.175................0.6

LTL : 0.175/1 < 0.6/1 

=> CH3COOH dư 

mCH3COOH (dư) = ( 0.6 - 0.175) * 60 = 25.5 (g) 

nCH3COOC2H5 = nC2H5OH = 0.175 (mol) 

H% = 0.14/0.175 * 100% = 80%

n tinh bột = 1,62/162n = 0,01/n(kmol)

$(C_6H_{10}O_5)_n + nH_2O \xrightarrow{t^o,xt} n C_6H_{12}O_6$

n glucozo = n . n tinh bột . H% = n . 0,01/n . 85% = 0,0085(kmol)

$C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH$

n C2H5OH = 2 . n glucozo . H% = 2.0,0085.90% = 0,0153(kmol)

$C_2H_5OH + O_2 \xrightarrow{t^o,xt} CH_3COOH + H_2O$

n CH3COOH = n C2H5OH .H% = 0,0153.70% = 0,01071(kmol)

m CH3COOH = 0,01071.60 = 0,6426(kg)

\(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{27,6}{46}=0,6\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO_4 đặc, t^o)

Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\), ta được C₂H₅OH dư nếu pư hết.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\)

\(\Rightarrow H=\dfrac{26,4}{44}.100\%=60\%\)