a) Khối lượng rượu etylic:

n_CO2 = \(\dfrac{11.2}{22.4}\) = 0.5 mol

Phương trình lên men glucozơ:

C6H12O6 → 2C2H5OH + 2CO2.

Theo pt: nrượu etylic = nCO2 = 0,5 mol.

mrượu etylic = 0,5 × 46 = 23g.

b) Khối lượng glucozơ.

Theo phương trình n_glucozo = 1/2. nCO2 = 1/2 . 0,5 = 0,25 mol

Do hiệu suất đạt 90% nên khối lượng glucozo cần dùng là:  0,25.180.\(\dfrac{100}{90}\)= 50 (g)

a) 

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: C₆H₁₂O₆ --men rượu--> 2C₂H₅OH + 2CO₂

              0,125<---------------------0,25<-------0,25

=> \(m_{C_2H_5OH}=0,25.46=11,5\left(g\right)\)

b) \(m_{C_6H_{12}O_6\left(pư\right)}=0,125.180=22,5\left(g\right)\)

=> \(m_{C_6H_{12}O_6\left(tt\right)}=\dfrac{22,5.100}{80}=28,125\left(g\right)\)

c) \(V_{C_2H_5OH}=\dfrac{11,5}{0,8}=14,375\left(ml\right)\)

=> \(V_{rượu}=\dfrac{14,375.100}{25}=57,5\left(ml\right)\)

C2hol+soc2😘😘😘

Nó gần như bao gồm cả bài nhưng cách giải chi tiết thì bn tự lm nha 😘♥️

\(n_{CO2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Pt : \(C_6H_{12}O_6\xrightarrow[30-35^oC]{Menrượu}2C_2H_5OH+2CO_2\)

                                           0,5            0,5

a) \(m_{C2H5OH}=0,5.46=23\left(g\right)\)

b) Pt : \(C_2H_5OH+O_2\xrightarrow[]{Mengiấm}CH_3COOH+H_2O\)

                0,5                                     0,5

\(m_{CH3COOH\left(lt\right)}=0,5.60=30\left(g\right)\)

⇒ \(m_{CH3COOH\left(tt\right)}=30.80\%=24\left(g\right)\)

 Chúc bạn học tốt

\(C_6H_{12}O_6\underrightarrow{t^o}2C_2H_5OH+2CO_2\uparrow\)(xt : men rượu )

                          0,5          0,5

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{C_2H_5OH}=0,5.46=23\left(g\right)\)

\(C_2H_5OH+O_2\underrightarrow{t^o}CH_3COOH+H_2O\) (men giấm )

0,5                                 0,5

\(m_{CH_3COOH}=0,5.60=30\left(g\right)\)

\(m_{CH_3COOHtt}=30.80\%=24\left(g\right)\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: \(C_6H_{12}O_6\dfrac{_{memruou}}{30^0-35^0C}>2C_2H_5OH+2CO_2\uparrow\)

               0,25                                  0,25              0,25      (mol)

a) \(m_{C_2H_5OH}=0,25.46=11,5\left(g\right)\)

b) \(m_{C_6H_{12}O_6}=0,25.180.90\%=40,5\left(g\right)\)

1)

a)

$C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$CH_3COOH + NaOH \to CH_3COONa + H_2O$

b)

$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

2)

a) $n_{CO_2} = \dfrac{16,8}{22,4} = 0,75(mol)$
$C_6H_{12}O_6 \xrightarrow{men\ rượu} 2CO_2 + 2C_2H_5OH$
$n_{glucozo} = \dfrac{1}{2}n_{CO_2} = 0,375(mol)$
$m_{glucozo} = 0,375.180 = 67,5(gam)$

b) $n_{C_2H_5OH} = n_{CO_2} = 0,75(mol)$
$m_{C_2H_5OH} = 0,75.46 = 34,5(gam)$

$V_{C_2H_5OH} = \dfrac{34,5}{0,8}=  43,125(ml)$

Câu 1:

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

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

\(CH_3COOH+Na\rightarrow CH_3COOH+\dfrac{1}{2}H_2\)

b, \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

Câu 2:

a, \(n_{CO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)

Theo PT: \(n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,375\left(mol\right)\)

\(\Rightarrow m_{C_6H_{12}O_6}=0,375.180=67,5\left(g\right)\)

b, \(n_{C_2H_5OH}=n_{CO_2}=0,75\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,75.46=34,5\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)