Bài 1:

PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)

\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)

Bài 2:

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết

\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)

Đáp án B

V_etylic = 460.8/100 = 36,8 ml => V_H2O = 423,2 ml =>m_H2O = 423,2g

=> m_etylic = 36,8.0,8 = 29,44g => n_etylic = 0,64 mol

C₂H₅OH +  O₂ → CH₃COOH + H₂O

=> n_CH3COOH = n_etylic.H% = 0,192 mol = n_O2

=> C%_CH3COOH = 0,192.60/(423,3 + 29,44 + 0,192.32) . 100% = 2,51%

a, \(n_{C_2H_4}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

PT: \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,7\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=0,7.90\%=0,63\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,63.46=28,98\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{28,98}{0,8}=36,225\left(ml\right)\)

b, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{C_2H_5OH}=0,63\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,63.60=37,8\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{37,8}{5\%}=756\left(g\right)\)

Đáp án A

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 D

Chọn đáp án D

Các phản ứng hóa học xảy ra:

Ta có

V_{Rượu nguyên chất} = 10 × 0,46 = 4,6 lít.

⇒ m_Rượu = V × D = 4,6 × 0.8 = 3,68 kg

⇒ n_C2H5OH = 0,08 kmol.

⇒ n_{Tinh bột} = 0,08 ÷ 2 ÷ 0,8 = 0,05 mol

⇒ m_{Tinh bột} = 8,1 kg

⇒ m_Gạo = 8,1 ÷ 0,75 = 10,8 kg

Đổi 10kg = 10000g

Ta có: \(n_{CH_3COOH\left(LT\right)}=\dfrac{10000.5\%}{92\%}=\dfrac{12500}{23}\left(mol\right)\)

PTHH:
\(C_2H_5OH+O_2\xrightarrow[]{\text{men giấm}}CH_3COOH+H_2O\)

\(\dfrac{12500}{23}\)<---------------------\(\dfrac{12500}{23}\)

\(\Rightarrow m_{C_2H_5OH}=\dfrac{12500}{23}.46=25000\left(g\right)=25\left(kg\right)\)